c F&A

The questions answered here are divided into several categories:

	 1. Declarations and Initializations
	 2. Structures, Unions, and Enumerations
	 3. Expressions
	 4. Pointers
	 5. Null Pointers
	 6. Arrays and Pointers
	 7. Memory Allocation
	 8. Characters and Strings
	 9. Boolean Expressions and Variables
	10. C Preprocessor
	11. ANSI/ISO Standard C
	12. Stdio
	13. Library Functions
	14. Floating Point
	15. Variable-Length Argument Lists
	16. Strange Problems
	17. Style
	18. Tools and Resources
	19. System Dependencies
	20. Miscellaneous
	    Bibliography
	    Acknowledgements

(The question numbers within each section are not always continuous,
because they are aligned with the aforementioned book-length version,
which contains even more questions.)

Herewith, some frequently-asked questions and their answers:


Section 1. Declarations and Initializations

1.1:	How should I decide which integer type to use?

A:	If you might need large values (above 32,767 or below -32,767),
	use long.  Otherwise, if space is very important (i.e. if there
	are large arrays or many structures), use short.  Otherwise, use
	int.  If well-defined overflow characteristics are important and
	negative values are not, or if you want to steer clear of sign-
	extension problems when manipulating bits or bytes, use one of
	the corresponding unsigned types.  (Beware when mixing signed
	and unsigned values in expressions, though.)

	Although character types (especially unsigned char) can be used
	as "tiny" integers, doing so is sometimes more trouble than it's
	worth, due to unpredictable sign extension and increased code
	size.  (Using unsigned char can help; see question 12.1 for a
	related problem.)

	A similar space/time tradeoff applies when deciding between
	float and double.  None of the above rules apply if pointers to
	the variable must have a particular type.

	If for some reason you need to declare something with an *exact*
	size (usually the only good reason for doing so is when
	attempting to conform to some externally-imposed storage layout,
	but see question 20.5), be sure to encapsulate the choice behind
	an appropriate typedef, such as those in C99's <inttypes.h>.

	If you need to manipulate huge values, larger than the
	guaranteed range of C's built-in types, see question 18.15d.

	References: K&R1 Sec. 2.2 p. 34; K&R2 Sec. 2.2 p. 36, Sec. A4.2
	pp. 195-6, Sec. B11 p. 257; ISO Sec. 5.2.4.2.1, Sec. 6.1.2.5;
	H&S Secs. 5.1,5.2 pp. 110-114.

1.4:	What should the 64-bit type be on a machine that can support it?

A:	The new C99 Standard specifies type long long as effectively
	being at least 64 bits, and this type has been implemented by a
	number of compilers for some time.  (Others have implemented
	extensions such as __longlong.)  On the other hand, it's also
	appropriate to implement type short int as 16, int as 32, and
	long int as 64 bits, and some compilers do.

	See also question 18.15d.

	References: C9X Sec. 5.2.4.2.1, Sec. 6.1.2.5.

1.7:	What's the best way to declare and define global variables
	and functions?

A:	First, though there can be many "declarations" (and in many
	translation units) of a single global variable or function,
	there must be exactly one "definition", where the definition is
	the declaration that actually allocates space, and provides an
	initialization value, if any.  The best arrangement is to place
	each definition in some relevant .c file, with an external
	declaration in a header (".h") file, which is included wherever
	the declaration is needed.  The .c file containing the
	definition should also #include the same header file, so the
	compiler can check that the definition matches the declarations.

	This rule promotes a high degree of portability: it is
	consistent with the requirements of the ANSI C Standard, and is
	also consistent with most pre-ANSI compilers and linkers.  (Unix
	compilers and linkers typically use a "common model" which
	allows multiple definitions, as long as at most one is
	initialized; this behavior is mentioned as a "common extension"
	by the ANSI Standard, no pun intended.)

	It is possible to use preprocessor tricks to arrange that a line
	like

		DEFINE(int, i);

	need only be entered once in one header file, and turned into a
	definition or a declaration depending on the setting of some
	macro, but it's not clear if this is worth the trouble.

	It's especially important to put global declarations in header
	files if you want the compiler to catch inconsistent
	declarations for you.  In particular, never place a prototype
	for an external function in a .c file: it wouldn't generally be
	checked for consistency with the definition, and an incompatible
	prototype is worse than useless.

	See also questions 10.6 and 18.8.

	References: K&R1 Sec. 4.5 pp. 76-7; K&R2 Sec. 4.4 pp. 80-1; ISO
	Sec. 6.1.2.2, Sec. 6.7, Sec. 6.7.2, Sec. G.5.11; Rationale
	Sec. 3.1.2.2; H&S Sec. 4.8 pp. 101-104, Sec. 9.2.3 p. 267; CT&P
	Sec. 4.2 pp. 54-56.

1.11:	What does extern mean in a function declaration?

A:	It can be used as a stylistic hint to indicate that the
	function's definition is probably in another source file, but
	there is no formal difference between

		extern int f();

	and

		int f();

	References: ISO Sec. 6.1.2.2, Sec. 6.5.1; Rationale
	Sec. 3.1.2.2; H&S Secs. 4.3,4.3.1 pp. 75-6.

1.12:	What's the auto keyword good for?

A:	Nothing; it's archaic.  See also question 20.37.

	References: K&R1 Sec. A8.1 p. 193; ISO Sec. 6.1.2.4, Sec. 6.5.1;
	H&S Sec. 4.3 p. 75, Sec. 4.3.1 p. 76.

1.14:	I can't seem to define a linked list successfully.  I tried

		typedef struct {
			char *item;
			NODEPTR next;
		} *NODEPTR;

	but the compiler gave me error messages.  Can't a structure in C
	contain a pointer to itself?

A:	Structures in C can certainly contain pointers to themselves;
	the discussion and example in section 6.5 of K&R make this
	clear.  The problem with the NODEPTR example is that the typedef
	has not yet been defined at the point where the "next" field is
	declared.  To fix this code, first give the structure a tag
	(e.g. "struct node").  Then, declare the "next" field as a
	simple "struct node *", or disentangle the typedef declaration
	from the structure definition, or both.  One corrected version
	would be

		struct node {
			char *item;
			struct node *next;
		};

		typedef struct node *NODEPTR;

	and there are at least three other equivalently correct ways of
	arranging it.

	A similar problem, with a similar solution, can arise when
	attempting to declare a pair of typedef'ed mutually referential
	structures.

	See also question 2.1.

	References: K&R1 Sec. 6.5 p. 101; K&R2 Sec. 6.5 p. 139; ISO
	Sec. 6.5.2, Sec. 6.5.2.3; H&S Sec. 5.6.1 pp. 132-3.

1.21:	How do I construct and understand declarations of complicated
	types such as "array of N pointers to functions returning
	pointers to functions returning pointers to char"?

A:	There are at least three ways of answering this question:

	1.  char *(*(*a[N])())();

	2.  Build the declaration up incrementally, using typedefs:

		typedef char *pc;	/* pointer to char */
		typedef pc fpc();	/* function returning pointer to char */
		typedef fpc *pfpc;	/* pointer to above */
		typedef pfpc fpfpc();	/* function returning... */
		typedef fpfpc *pfpfpc;	/* pointer to... */
		pfpfpc a[N];		/* array of... */

	3.  Use the cdecl program, which turns English into C and vice
	    versa:

		cdecl> declare a as array of pointer to function returning
			pointer to function returning pointer to char
		char *(*(*a[])())()

	    cdecl can also explain complicated declarations, help with
	    casts, and indicate which set of parentheses the parameters
	    go in (for complicated function definitions, like the one
	    above).  See question 18.1.

	A good book on C should explain how to read these complicated
	declarations "inside out" to understand them ("declaration
	mimics use").

	The pointer-to-function declarations in the examples above have
	not included parameter type information.  When the parameters
	have complicated types, declarations can *really* get messy.
	(Modern versions of cdecl can help here, too.)

	References: K&R2 Sec. 5.12 p. 122; ISO Sec. 6.5ff (esp.
	Sec. 6.5.4); H&S Sec. 4.5 pp. 85-92, Sec. 5.10.1 pp. 149-50.

1.25:	My compiler is complaining about an invalid redeclaration of a
	function, but I only define it once and call it once.

A:	Functions which are called without a declaration in scope,
	perhaps because the first call precedes the function's
	definition, are assumed to be declared as returning int (and
	without any argument type information), leading to discrepancies
	if the function is later declared or defined otherwise.  All
	functions should be (and non-int functions must be) declared
	before they are called.

	Another possible source of this problem is that the function has
	the same name as another one declared in some header file.

	See also questions 11.3 and 15.1.

	References: K&R1 Sec. 4.2 p. 70; K&R2 Sec. 4.2 p. 72; ISO
	Sec. 6.3.2.2; H&S Sec. 4.7 p. 101.

1.25b:	What's the right declaration for main()?
	Is void main() correct?

A:	See questions 11.12a through 11.15.  (But no, it's not correct.)

1.30:	What am I allowed to assume about the initial values of
	variables and arrays which are not explicitly initialized?
	If global variables start out as "zero", is that good enough
	for null pointers and floating-point zeroes?

A:	Uninitialized variables with "static" duration (that is, those
	declared outside of functions, and those declared with the
	storage class static), are guaranteed to start out as zero, just
	as if the programmer had typed "= 0".  Therefore, such variables
	are implicitly initialized to the null pointer (of the correct
	type; see also section 5) if they are pointers, and to 0.0 if
	they are floating-point.

	Variables with "automatic" duration (i.e. local variables
	without the static storage class) start out containing garbage,
	unless they are explicitly initialized.  (Nothing useful can be
	predicted about the garbage.)

	These rules do apply to arrays and structures (termed
	"aggregates"); arrays and structures are considered "variables"
	as far as initialization is concerned.

	Dynamically-allocated memory obtained with malloc() and
	realloc() is likely to contain garbage, and must be initialized
	by the calling program, as appropriate.  Memory obtained with
	calloc() is all-bits-0, but this is not necessarily useful for
	pointer or floating-point values (see question 7.31, and section
	5).

	References: K&R1 Sec. 4.9 pp. 82-4; K&R2 Sec. 4.9 pp. 85-86; ISO
	Sec. 6.5.7, Sec. 7.10.3.1, Sec. 7.10.5.3; H&S Sec. 4.2.8 pp.
	72-3, Sec. 4.6 pp. 92-3, Sec. 4.6.2 pp. 94-5, Sec. 4.6.3 p. 96,
	Sec. 16.1 p. 386.

1.31:	This code, straight out of a book, isn't compiling:

		int f()
		{
			char a[] = "Hello, world!";
		}

A:	Perhaps you have an old, pre-ANSI compiler, which doesn't allow
	initialization of "automatic aggregates" (i.e. non-static local
	arrays, structures, or unions).  See also question 11.29.

1.31b:	What's wrong with this initialization?

		char *p = malloc(10);

	My compiler is complaining about an "invalid initializer",
	or something.

A:	Is the declaration of a static or non-local variable?  Function
	calls are allowed in initializers only for automatic variables
	(that is, for local, non-static variables).

1.32:	What is the difference between these initializations?

		char a[] = "string literal";
		char *p  = "string literal";

	My program crashes if I try to assign a new value to p[i].

A:	A string literal can be used in two slightly different ways.  As
	an array initializer (as in the declaration of char a[] in the
	question), it specifies the initial values of the characters in
	that array.  Anywhere else, it turns into an unnamed, static
	array of characters, which may be stored in read-only memory,
	and which therefore cannot necessarily be modified.  In an
	expression context, the array is converted at once to a pointer,
	as usual (see section 6), so the second declaration initializes
	p to point to the unnamed array's first element.

	(For compiling old code, some compilers have a switch
	controlling whether string literals are writable or not.)

	See also questions 1.31, 6.1, 6.2, 6.8, and 11.8b.

	References: K&R2 Sec. 5.5 p. 104; ISO Sec. 6.1.4, Sec. 6.5.7;
	Rationale Sec. 3.1.4; H&S Sec. 2.7.4 pp. 31-2.

1.34:	I finally figured out the syntax for declaring pointers to
	functions, but now how do I initialize one?

A:	Use something like

		extern int func();
		int (*fp)() = func;

	When the name of a function appears in an expression, it
	"decays" into a pointer (that is, it has its address implicitly
	taken), much as an array name does.

	A prior, explicit declaration for the function (perhaps in a
	header file) is normally needed.  The implicit external function
	declaration that can occur when a function is called does not
	help when a function name's only use is for its value.

	See also questions 1.25 and 4.12.


Section 2. Structures, Unions, and Enumerations

2.1:	What's the difference between these two declarations?

		struct x1 { ... };
		typedef struct { ... } x2;

A:	The first form declares a "structure tag"; the second declares a
	"typedef".  The main difference is that you subsequently refer
	to the first type as "struct x1" and the second simply as "x2".
	That is, the second declaration is of a slightly more abstract
	type -- its users don't necessarily know that it is a structure,
	and the keyword struct is not used when declaring instances of it.

2.2:	Why doesn't

		struct x { ... };
		x thestruct;

	work?

A:	C is not C++.  Typedef names are not automatically generated for
	structure tags.  See also questions 1.14 and 2.1.

2.3:	Can a structure contain a pointer to itself?

A:	Most certainly.  See also question 1.14.

2.4:	How can I implement opaque (abstract) data types in C?

A:	One good way is for clients to use structure pointers (perhaps
	additionally hidden behind typedefs) which point to structure
	types which are not publicly defined.  It's legal to declare
	and use "anonymous" structure pointers (that is, pointers to
	structures of incomplete type), as long as no attempt is made to
	access the members -- which of course is exactly the point of an
	opaque type.

2.4b:	Is there a good way of simulating OOP-style inheritance, or
	other OOP features, in C?

A:	It's straightforward to implement simple "methods" by placing
	function pointers in structures.  You can make various clumsy,
	brute-force attempts at inheritance using the preprocessor or by
	having structures contain "base types" as initial subsets, but
	it won't be perfect.  There's obviously no operator overloading,
	and overriding (i.e. of "methods" in "derived classes") would
	have to be done by hand.

	Obviously, if you need "real" OOP, you'll want to use a language
	that supports it, such as C++.

2.6:	I came across some code that declared a structure like this:

		struct name {
			int namelen;
			char namestr[1];
		};

	and then did some tricky allocation to make the namestr array
	act like it had several elements.  Is this legal or portable?

A:	This technique is popular, although Dennis Ritchie has called it
	"unwarranted chumminess with the C implementation."  An official
	interpretation has deemed that it is not strictly conforming
	with the C Standard, although it does seem to work under all
	known implementations.  (Compilers which check array bounds
	carefully might issue warnings.)

	Another possibility is to declare the variable-size element very
	large, rather than very small; in the case of the above example:

		...
		char namestr[MAXSIZE];

	where MAXSIZE is larger than any name which will be stored.
	However, it looks like this technique is disallowed by a strict
	interpretation of the Standard as well.  Furthermore, either of
	these "chummy" structures must be used with care, since the
	programmer knows more about their size than the compiler does.

	C99 introduces the concept of a "flexible array member", which
	allows the size of an array to be omitted if it is the last
	member in a structure, thus providing a well-defined solution.

	References: Rationale Sec. 3.5.4.2; C9X Sec. 6.5.2.1.

2.8:	Is there a way to compare structures automatically?

A:	No.  There is not a good way for a compiler to implement
	structure comparison (i.e. to support the == operator for
	structures) which is consistent with C's low-level flavor.
	A simple byte-by-byte comparison could founder on random bits
	present in unused "holes" in the structure (see question 2.12).
	A field-by-field comparison might require unacceptable amounts
	of repetitive code for large structures.

	If you need to compare two structures, you'll have to write your
	own function to do so, field by field.

	References: K&R2 Sec. 6.2 p. 129; Rationale Sec. 3.3.9; H&S
	Sec. 5.6.2 p. 133.

2.10:	How can I pass constant values to functions which accept
	structure arguments?

A:	Traditional C had no way of generating anonymous structure
	values; you had to use a temporary structure variable or a
	little structure-building function.

	C99 introduces "compound literals", one form of which provides
	for structure constants.  For example, to pass a constant
	coordinate pair to a hypothetical plotpoint() function which
	expects a struct point, you can call

		plotpoint((struct point){1, 2});

	Combined with "designated initializers" (another C99 feature),
	it is also possible to specify member values by name:

		plotpoint((struct point){.x=1, .y=2});

	See also question 4.10.

	References: C9X Sec. 6.3.2.5, Sec. 6.5.8.

2.11:	How can I read/write structures from/to data files?

A:	It is relatively straightforward to write a structure out using
	fwrite():

		fwrite(&somestruct, sizeof somestruct, 1, fp);

	and a corresponding fread invocation can read it back in.
	However, data files so written will *not* be portable (see
	questions 2.12 and 20.5).  Also, if the structure contains any
	pointers, only the pointer values will be written, and they are
	most unlikely to be valid when read back in.  Finally, note that
	for widespread portability you must use the "b" flag when
	opening the files; see question 12.38.

	A more portable solution, though it's a bit more work initially,
	is to write a pair of functions for writing and reading a
	structure, field-by-field, in a portable (perhaps even human-
	readable) way.

	References: H&S Sec. 15.13 p. 381.

2.12:	My compiler is leaving holes in structures, which is wasting
	space and preventing "binary" I/O to external data files.  Why?
	Can I turn this off, or otherwise control the alignment of
	structure fields?

A:	Those "holes" provide "padding", which may be needed in order to
	preserve the "alignment" of later fields of the structure.  For
	efficient access, most processors prefer (or require) that
	multibyte objects (e.g. structure members of any type larger
	than char) not sit at arbitrary memory addresses, but rather at
	addresses which are multiples of 2 or 4 or the object size.

	Your compiler may provide an extension to give you explicit
	control over struct alignment (perhaps involving a #pragma; see
	question 11.20), but there is no standard method.

	See also question 20.5.

	References: K&R2 Sec. 6.4 p. 138; H&S Sec. 5.6.4 p. 135.

2.13:	Why does sizeof report a larger size than I expect for a
	structure type, as if there were padding at the end?

A:	Padding at the end of a structure may be necessary to preserve
	alignment when an array of contiguous structures is allocated.
	Even when the structure is not part of an array, the padding
	remains, so that sizeof can always return a consistent size.
	See also question 2.12 above.

	References: H&S Sec. 5.6.7 pp. 139-40.

2.14:	How can I determine the byte offset of a field within a
	structure?

A:	ANSI C defines the offsetof() macro in <stddef.h>, which lets
	you compute the offset of field f in struct s as
	offsetof(struct s, f).  If for some reason you have to code this
	sort of thing yourself, one possibility is

		#define offsetof(type, f) ((size_t) \
			((char *)&((type *)0)->f - (char *)(type *)0))

	This implementation is not 100% portable; some compilers may
	legitimately refuse to accept it.

	References: ISO Sec. 7.1.6; Rationale Sec. 3.5.4.2; H&S
	Sec. 11.1 pp. 292-3.

2.15:	How can I access structure fields by name at run time?

A:	Keep track of the field offsets as computed using the offsetof()
	macro (see question 2.14).  If structp is a pointer to an
	instance of the structure, and field f is an int having offset
	offsetf, f's value can be set indirectly with

		*(int *)((char *)structp + offsetf) = value;

2.18:	This program works correctly, but it dumps core after it
	finishes.  Why?

		struct list {
			char *item;
			struct list *next;
		}

		/* Here is the main program. */

		main(argc, argv)
		{ ... }

A:	A missing semicolon causes main() to be declared as returning a
	structure.  (The connection is hard to see because of the
	intervening comment.)  Since structure-valued functions are
	usually implemented by adding a hidden return pointer, the
	generated code for main() tries to accept three arguments,
	although only two are passed (in this case, by the C start-up
	code).  See also questions 10.9 and 16.4.

	References: CT&P Sec. 2.3 pp. 21-2.

2.20:	Can I initialize unions?

A:	In the original ANSI C, an initializer was allowed only for the
	first-named member of a union.  C99 introduces "designated
	initializers" which can be used to initialize any member.

	References: K&R2 Sec. 6.8 pp. 148-9; ISO Sec. 6.5.7; C9X
	Sec. 6.5.8; H&S Sec. 4.6.7 p. 100.

2.22:	What's the difference between an enumeration and a set of
	preprocessor #defines?

A:	There is little difference.  The C Standard says that
	enumerations may be freely intermixed with other integral types,
	without errors.  (If, on the other hand, such intermixing were
	disallowed without explicit casts, judicious use of enumerations
	could catch certain programming errors.)

	Some advantages of enumerations are that the numeric values are
	automatically assigned, that a debugger may be able to display
	the symbolic values when enumeration variables are examined, and
	that they obey block scope.  (A compiler may also generate
	nonfatal warnings when enumerations are indiscriminately mixed,
	since doing so can still be considered bad style.)  A
	disadvantage is that the programmer has little control over
	those nonfatal warnings; some programmers also resent not having
	control over the sizes of enumeration variables.

	References: K&R2 Sec. 2.3 p. 39, Sec. A4.2 p. 196; ISO
	Sec. 6.1.2.5, Sec. 6.5.2, Sec. 6.5.2.2, Annex F; H&S Sec. 5.5
	pp. 127-9, Sec. 5.11.2 p. 153.

2.24:	Is there an easy way to print enumeration values symbolically?

A:	No.  You can write a little function to map an enumeration
	constant to a string.  (For debugging purposes, a good debugger
	should automatically print enumeration constants symbolically.)


Section 3. Expressions

3.1:	Why doesn't this code:

		a[i] = i++;

	work?

A:	The subexpression i++ causes a side effect -- it modifies i's
	value -- which leads to undefined behavior since i is also
	referenced elsewhere in the same expression, and there's no way
	to determine whether the reference (in a[i] on the left-hand
	side) should be to the old or the new value.  (Note that
	although the language in K&R suggests that the behavior of this
	expression is unspecified, the C Standard makes the stronger
	statement that it is undefined -- see question 11.33.)

	References: K&R1 Sec. 2.12; K&R2 Sec. 2.12; ISO Sec. 6.3; H&S
	Sec. 7.12 pp. 227-9.

3.2:	Under my compiler, the code

		int i = 7;
		printf("%d\n", i++ * i++);

	prints 49.  Regardless of the order of evaluation, shouldn't it
	print 56?

A:	Although the postincrement and postdecrement operators ++ and --
	perform their operations after yielding the former value, the
	implication of "after" is often misunderstood.  It is *not*
	guaranteed that an increment or decrement is performed
	immediately after giving up the previous value and before any
	other part of the expression is evaluated.  It is merely
	guaranteed that the update will be performed sometime before the
	expression is considered "finished" (before the next "sequence
	point," in ANSI C's terminology; see question 3.8).  In the
	example, the compiler chose to multiply the previous value by
	itself and to perform both increments later.

	The behavior of code which contains multiple, ambiguous side
	effects has always been undefined.  (Loosely speaking, by
	"multiple, ambiguous side effects" we mean any combination of
	increment, decrement, and assignment operators in a single
	expression which causes the same object either to be modified
	twice or modified and then inspected.  This is a rough
	definition; see question 3.8 for a precise one, and question
	11.33 for the meaning of "undefined.")  Don't even try to find
	out how your compiler implements such things (contrary to the
	ill-advised exercises in many C textbooks); as K&R wisely point
	out, "if you don't know *how* they are done on various machines,
	that innocence may help to protect you."

	References: K&R1 Sec. 2.12 p. 50; K&R2 Sec. 2.12 p. 54; ISO
	Sec. 6.3; H&S Sec. 7.12 pp. 227-9; CT&P Sec. 3.7 p. 47; PCS
	Sec. 9.5 pp. 120-1.

3.3:	I've experimented with the code

		int i = 3;
		i = i++;

	on several compilers.  Some gave i the value 3, and some gave 4.
	Which compiler is correct?

A:	There is no correct answer; the expression is undefined.  See
	questions 3.1, 3.8, 3.9, and 11.33.  (Also, note that neither
	i++ nor ++i is the same as i+1.  If you want to increment i,
	use i=i+1, i+=1, i++, or ++i, not some combination.  See also
	question 3.12b.)

3.3b:	Here's a slick expression:

		a ^= b ^= a ^= b

	It swaps a and b without using a temporary.

A:	Not portably, it doesn't.  It attempts to modify the variable a
	twice between sequence points, so its behavior is undefined.

	For example, it has been reported that when given the code

		int a = 123, b = 7654;
		a ^= b ^= a ^= b;

	the SCO Optimizing C compiler (icc) sets b to 123 and a to 0.

	See also questions 3.1, 3.8, 10.3, and 20.15c.

3.4:	Can I use explicit parentheses to force the order of evaluation
	I want?  Even if I don't, doesn't precedence dictate it?

A:	Not in general.

	Operator precedence and explicit parentheses impose only a
	partial ordering on the evaluation of an expression.  In the
	expression

		f() + g() * h()

	although we know that the multiplication will happen before the
	addition, there is no telling which of the three functions will
	be called first.

	When you need to ensure the order of subexpression evaluation,
	you may need to use explicit temporary variables and separate
	statements.

	References: K&R1 Sec. 2.12 p. 49, Sec. A.7 p. 185; K&R2
	Sec. 2.12 pp. 52-3, Sec. A.7 p. 200.

3.5:	But what about the && and || operators?
	I see code like "while((c = getchar()) != EOF && c != '\n')" ...

A:	There is a special "short-circuiting" exception for these
	operators: the right-hand side is not evaluated if the left-hand
	side determines the outcome (i.e. is true for || or false for
	&&).  Therefore, left-to-right evaluation is guaranteed, as it
	also is for the comma operator.  Furthermore, all of these
	operators (along with ?:) introduce an extra internal sequence
	point (see question 3.8).

	References: K&R1 Sec. 2.6 p. 38, Secs. A7.11-12 pp. 190-1; K&R2
	Sec. 2.6 p. 41, Secs. A7.14-15 pp. 207-8; ISO Sec. 6.3.13,
	Sec. 6.3.14, Sec. 6.3.15; H&S Sec. 7.7 pp. 217-8, Sec. 7.8 pp.
	218-20, Sec. 7.12.1 p. 229; CT&P Sec. 3.7 pp. 46-7.

3.8:	How can I understand these complex expressions?  What's a
	"sequence point"?

A:	A sequence point is a point in time (at the end of the
	evaluation of a full expression, or at the ||, &&, ?:, or comma
	operators, or just before a function call) at which the dust
	has settled and all side effects are guaranteed to be complete.
	The ANSI/ISO C Standard states that

		Between the previous and next sequence point an
		object shall have its stored value modified at
		most once by the evaluation of an expression.
		Furthermore, the prior value shall be accessed
		only to determine the value to be stored.

	The second sentence can be difficult to understand.  It says
	that if an object is written to within a full expression, any
	and all accesses to it within the same expression must be
	directly involved in the computation of the value to be written.
	This rule effectively constrains legal expressions to those in
	which the accesses demonstrably precede the modification.  For
	example, i = i + 1 is legal, but not a[i] = i++ (see question
	3.1).

	See also question 3.9 below.

	References: ISO Sec. 5.1.2.3, Sec. 6.3, Sec. 6.6, Annex C;
	Rationale Sec. 2.1.2.3; H&S Sec. 7.12.1 pp. 228-9.

3.9:	So given

		a[i] = i++;

	we don't know which cell of a[] gets written to, but i does get
	incremented by one, right?

A:	Not necessarily!  Once an expression or program becomes
	undefined, *all* aspects of it become undefined.  See questions
	3.2, 3.3, 11.33, and 11.35.

3.12a:	What's the difference between ++i and i++?

A:	If your C book doesn't explain, get a better one.  Briefly:
	++i adds one to the stored value of i and "returns" the new,
	incremented value to the surrounding expression; i++ adds one
	to i but returns the prior, unincremented value.

3.12b:	If I'm not using the value of the expression, should I use ++i
	or i++ to increment a variable?

A:	Since the two forms differ only in the value yielded, they are
	entirely equivalent when only their side effect is needed.
	(However, the prefix form is preferred in C++.)  See also
	question 3.3.

	References: K&R1 Sec. 2.8 p. 43; K&R2 Sec. 2.8 p. 47; ISO
	Sec. 6.3.2.4, Sec. 6.3.3.1; H&S Sec. 7.4.4 pp. 192-3, Sec. 7.5.8
	pp. 199-200.

3.14:	Why doesn't the code

		int a = 1000, b = 1000;
		long int c = a * b;

	work?

A:	Under C's integral promotion rules, the multiplication is
	carried out using int arithmetic, and the result may overflow or
	be truncated before being promoted and assigned to the long int
	left-hand side.  Use an explicit cast to force long arithmetic:

		long int c = (long int)a * b;

	Notice that (long int)(a * b) would *not* have the desired
	effect.

	A similar problem can arise when two integers are divided, with
	the result assigned to a floating-point variable; the solution
	is similar, too.

	References: K&R1 Sec. 2.7 p. 41; K&R2 Sec. 2.7 p. 44; ISO
	Sec. 6.2.1.5; H&S Sec. 6.3.4 p. 176; CT&P Sec. 3.9 pp. 49-50.

3.16:	I have a complicated expression which I have to assign to one of
	two variables, depending on a condition.  Can I use code like
	this?

		((condition) ? a : b) = complicated_expression;

A:	No.  The ?: operator, like most operators, yields a value, and
	you can't assign to a value.  (In other words, ?: does not yield
	an "lvalue".)  If you really want to, you can try something like

		*((condition) ? &a : &b) = complicated_expression;

	although this is admittedly not as pretty.

	References: ISO Sec. 6.3.15; H&S Sec. 7.1 pp. 179-180.


Section 4. Pointers

4.2:	I'm trying to declare a pointer and allocate some space for it,
	but it's not working.  What's wrong with this code?

		char *p;
		*p = malloc(10);

A:	The pointer you declared is p, not *p.  When you're manipulating
	the pointer itself (for example when you're setting it to make
	it point somewhere), you just use the name of the pointer:

		p = malloc(10);

	It's when you're manipulating the pointed-to memory that you use
	* as an indirection operator:

		*p = 'H';

	See also questions 1.21, 7.1, 7.3c, and 8.3.

	References: CT&P Sec. 3.1 p. 28.

4.3:	Does *p++ increment p, or what it points to?

A:	The postfix ++ and -- operators essentially have higher
	precedence than the prefix unary operators.  Therefore, *p++ is
	equivalent to *(p++); it increments p, and returns the value
	which p pointed to before p was incremented.  To increment the
	value pointed to by p, use (*p)++ (or perhaps ++*p, if the order
	of the side effect doesn't matter).

	References: K&R1 Sec. 5.1 p. 91; K&R2 Sec. 5.1 p. 95; ISO
	Sec. 6.3.2, Sec. 6.3.3; H&S Sec. 7.4.4 pp. 192-3, Sec. 7.5 p.
	193, Secs. 7.5.7,7.5.8 pp. 199-200.

4.5:	I have a char * pointer that happens to point to some ints, and
	I want to step it over them.  Why doesn't

		((int *)p)++;

	work?

A:	In C, a cast operator does not mean "pretend these bits have a
	different type, and treat them accordingly"; it is a conversion
	operator, and by definition it yields an rvalue, which cannot be
	assigned to, or incremented with ++.  (It is either an accident
	or a deliberate but nonstandard extension if a particular
	compiler accepts expressions such as the above.)  Say what you
	mean: use

		p = (char *)((int *)p + 1);

	or (since p is a char *) simply

		p += sizeof(int);

	When possible, however, you should choose appropriate pointer
	types in the first place, rather than trying to treat one type
	as another.

	References: K&R2 Sec. A7.5 p. 205; ISO Sec. 6.3.4; Rationale
	Sec. 3.3.2.4; H&S Sec. 7.1 pp. 179-80.

4.8:	I have a function which accepts, and is supposed to initialize,
	a pointer:

		void f(int *ip)
		{
			static int dummy = 5;
			ip = &dummy;
		}

	But when I call it like this:

		int *ip;
		f(ip);

	the pointer in the caller remains unchanged.

A:	Are you sure the function initialized what you thought it did?
	Remember that arguments in C are passed by value.  The called
	function altered only the passed copy of the pointer.  You'll
	either want to pass the address of the pointer (the function
	will end up accepting a pointer-to-a-pointer), or have the
	function return the pointer.

	See also questions 4.9 and 4.11.

4.9:	Can I use a void ** pointer as a parameter so that a function
	can accept a generic pointer by reference?

A:	Not portably.  There is no generic pointer-to-pointer type in C.
	void * acts as a generic pointer only because conversions (if
	necessary) are applied automatically when other pointer types
	are assigned to and from void *'s; these conversions cannot be
	performed (the correct underlying pointer type is not known) if
	an attempt is made to indirect upon a void ** value which points
	at a pointer type other than void *.

4.10:	I have a function

		extern int f(int *);

	which accepts a pointer to an int.  How can I pass a constant by
	reference?  A call like

		f(&5);

	doesn't seem to work.

A:	In C99, you can use a "compound literal":

		f((int[]){5});

	Prior to C99, you couldn't do this directly; you had to declare
	a temporary variable, and then pass its address to the function:

		int five = 5;
		f(&five);

	See also questions 2.10, 4.8, and 20.1.

4.11:	Does C even have "pass by reference"?

A:	Not really.

	Strictly speaking, C always uses pass by value.  You can
	simulate pass by reference yourself, by defining functions which
	accept pointers and then using the & operator when calling, and
	the compiler will essentially simulate it for you when you pass
	an array to a function (by passing a pointer instead, see
	question 6.4 et al.).  However, C has nothing truly equivalent
	to formal pass by reference or C++ reference parameters.  (On
	the other hand, function-like preprocessor macros can provide a
	form of "pass by name".)

	See also questions 4.8 and 20.1.

	References: K&R1 Sec. 1.8 pp. 24-5, Sec. 5.2 pp. 91-3; K&R2
	Sec. 1.8 pp. 27-8, Sec. 5.2 pp. 95-7; ISO Sec. 6.3.2.2; H&S
	Sec. 9.5 pp. 273-4.

4.12:	I've seen different syntax used for calling functions via
	pointers.  What's the story?

A:	Originally, a pointer to a function had to be "turned into" a
	"real" function, with the * operator (and an extra pair of
	parentheses, to keep the precedence straight), before calling:

		int r, func(), (*fp)() = func;
		r = (*fp)();

	It can also be argued that functions are always called via
	pointers, and that "real" function names always decay implicitly
	into pointers (in expressions, as they do in initializations;
	see question 1.34).  This reasoning means that

		r = fp();

	is legal and works correctly, whether fp is the name of a
	function or a pointer to one.  (The usage has always been
	unambiguous; there is nothing you ever could have done with a
	function pointer followed by an argument list except call the
	function pointed to.)

	The ANSI C Standard essentially adopts the latter
	interpretation, meaning that the explicit * is not required,
	though it is still allowed.

	See also question 1.34.

	References: K&R1 Sec. 5.12 p. 116; K&R2 Sec. 5.11 p. 120; ISO
	Sec. 6.3.2.2; Rationale Sec. 3.3.2.2; H&S Sec. 5.8 p. 147,
	Sec. 7.4.3 p. 190.

4.15:	How do I convert an int to a char *?  I tried a cast, but it's
	not working.

A:	It depends on what you're trying to do.  If you tried a cast
	but it's not working, you're probably trying to convert an
	integer to a string, in which case see question 13.1.  If you're
	trying to convert an integer to a character, see question 8.6.
	If you're trying to set a pointer to point to a particular
	memory address, see question 19.25.


Section 5. Null Pointers

5.1:	What is this infamous null pointer, anyway?

A:	The language definition states that for each pointer type, there
	is a special value -- the "null pointer" -- which is
	distinguishable from all other pointer values and which is
	"guaranteed to compare unequal to a pointer to any object or
	function."  That is, the address-of operator & will never yield
	a null pointer, nor will a successful call to malloc().
	(malloc() does return a null pointer when it fails, and this is
	a typical use of null pointers: as a "special" pointer value
	with some other meaning, usually "not allocated" or "not
	pointing anywhere yet.")

	A null pointer is conceptually different from an uninitialized
	pointer.  A null pointer is known not to point to any object or
	function; an uninitialized pointer might point anywhere.  See
	also questions 1.30, 7.1, and 7.31.

	As mentioned above, there is a null pointer for each pointer
	type, and the internal values of null pointers for different
	types may be different.  Although programmers need not know the
	internal values, the compiler must always be informed which type
	of null pointer is required, so that it can make the distinction
	if necessary (see questions 5.2, 5.5, and 5.6 below).

	References: K&R1 Sec. 5.4 pp. 97-8; K&R2 Sec. 5.4 p. 102; ISO
	Sec. 6.2.2.3; Rationale Sec. 3.2.2.3; H&S Sec. 5.3.2 pp. 121-3.

5.2:	How do I get a null pointer in my programs?

A:	According to the language definition, a constant 0 in a pointer
	context is converted into a null pointer at compile time.  That
	is, in an initialization, assignment, or comparison when one
	side is a variable or expression of pointer type, the compiler
	can tell that a constant 0 on the other side requests a null
	pointer, and generate the correctly-typed null pointer value.
	Therefore, the following fragments are perfectly legal:

		char *p = 0;
		if(p != 0)

	(See also question 5.3.)

	However, an argument being passed to a function is not
	necessarily recognizable as a pointer context, and the compiler
	may not be able to tell that an unadorned 0 "means" a null
	pointer.  To generate a null pointer in a function call context,
	an explicit cast may be required, to force the 0 to be
	recognized as a pointer.  For example, the Unix system call
	execl takes a variable-length, null-pointer-terminated list of
	character pointer arguments, and is correctly called like this:

		execl("/bin/sh", "sh", "-c", "date", (char *)0);

	If the (char *) cast on the last argument were omitted, the
	compiler would not know to pass a null pointer, and would pass
	an integer 0 instead.  (Note that many Unix manuals get this
	example wrong.)

	When function prototypes are in scope, argument passing becomes
	an "assignment context," and most casts may safely be omitted,
	since the prototype tells the compiler that a pointer is
	required, and of which type, enabling it to correctly convert an
	unadorned 0.  Function prototypes cannot provide the types for
	variable arguments in variable-length argument lists however, so
	explicit casts are still required for those arguments.  (See
	also question 15.3.)  It is probably safest to properly cast
	all null pointer constants in function calls, to guard against
	varargs functions or those without prototypes.

	Summary:

		Unadorned 0 okay:	Explicit cast required:

		initialization		function call,
					no prototype in scope
		assignment
					variable argument in
		comparison		varargs function call

		function call,
		prototype in scope,
		fixed argument

	References: K&R1 Sec. A7.7 p. 190, Sec. A7.14 p. 192; K&R2
	Sec. A7.10 p. 207, Sec. A7.17 p. 209; ISO Sec. 6.2.2.3; H&S
	Sec. 4.6.3 p. 95, Sec. 6.2.7 p. 171.

5.3:	Is the abbreviated pointer comparison "if(p)" to test for non-
	null pointers valid?  What if the internal representation for
	null pointers is nonzero?

A:	When C requires the Boolean value of an expression, a false
	value is inferred when the expression compares equal to zero,
	and a true value otherwise.  That is, whenever one writes

		if(expr)

	where "expr" is any expression at all, the compiler essentially
	acts as if it had been written as

		if((expr) != 0)

	Substituting the trivial pointer expression "p" for "expr", we
	have

		if(p)	is equivalent to		if(p != 0)

	and this is a comparison context, so the compiler can tell that
	the (implicit) 0 is actually a null pointer constant, and use
	the correct null pointer value.  There is no trickery involved
	here; compilers do work this way, and generate identical code
	for both constructs.  The internal representation of a null
	pointer does *not* matter.

	The boolean negation operator, !, can be described as follows:

		!expr	is essentially equivalent to	(expr)?0:1
			or to				((expr) == 0)

	which leads to the conclusion that

		if(!p)	is equivalent to		if(p == 0)

	"Abbreviations" such as if(p), though perfectly legal, are
	considered by some to be bad style (and by others to be good
	style; see question 17.10).

	See also question 9.2.

	References: K&R2 Sec. A7.4.7 p. 204; ISO Sec. 6.3.3.3,
	Sec. 6.3.9, Sec. 6.3.13, Sec. 6.3.14, Sec. 6.3.15, Sec. 6.6.4.1,
	Sec. 6.6.5; H&S Sec. 5.3.2 p. 122.

5.4:	What is NULL and how is it defined?

A:	As a matter of style, many programmers prefer not to have
	unadorned 0's scattered through their programs.  Therefore, the
	preprocessor macro NULL is defined (by <stdio.h> and several
	other headers) as a null pointer constant, typically 0 or
	((void *)0) (see also question 5.6).  A programmer who wishes to
	make explicit the distinction between 0 the integer and 0 the
	null pointer constant can then use NULL whenever a null pointer
	is required.

	Using NULL is a stylistic convention only; the preprocessor
	turns NULL back into 0 which is then recognized by the compiler,
	in pointer contexts, as before.  In particular, a cast may still
	be necessary before NULL (as before 0) in a function call
	argument.  The table under question 5.2 above applies for NULL
	as well as 0 (an unadorned NULL is equivalent to an unadorned
	0).

	NULL should be used *only* as a pointer constant; see question 5.9.

	References: K&R1 Sec. 5.4 pp. 97-8; K&R2 Sec. 5.4 p. 102; ISO
	Sec. 7.1.6, Sec. 6.2.2.3; Rationale Sec. 4.1.5; H&S Sec. 5.3.2
	p. 122, Sec. 11.1 p. 292.

5.5:	How should NULL be defined on a machine which uses a nonzero bit
	pattern as the internal representation of a null pointer?

A:	The same as on any other machine: as 0 (or some version of 0;
	see question 5.4).

	Whenever a programmer requests a null pointer, either by writing
	"0" or "NULL", it is the compiler's responsibility to generate
	whatever bit pattern the machine uses for that null pointer.
	Therefore, #defining NULL as 0 on a machine for which internal
	null pointers are nonzero is as valid as on any other: the
	compiler must always be able to generate the machine's correct
	null pointers in response to unadorned 0's seen in pointer
	contexts.  See also questions 5.2, 5.10, and 5.17.

	References: ISO Sec. 7.1.6; Rationale Sec. 4.1.5.

5.6:	If NULL were defined as follows:

		#define NULL ((char *)0)

	wouldn't that make function calls which pass an uncast NULL
	work?

A:	Not in the most general case.  The complication is that there
	are machines which use different internal representations for
	pointers to different types of data.  The suggested definition
	would make uncast NULL arguments to functions expecting pointers
	to characters work correctly, but pointer arguments of other
	types could still (in the absence of prototypes) be
	problematical, and legal constructions such as

		FILE *fp = NULL;

	could fail.

	Nevertheless, ANSI C allows the alternate definition

		#define NULL ((void *)0)

	for NULL.  Besides potentially helping incorrect programs to
	work (but only on machines with homogeneous pointers, thus
	questionably valid assistance), this definition may catch
	programs which use NULL incorrectly (e.g. when the ASCII NUL
	character was really intended; see question 5.9).

	At any rate, ANSI function prototypes ensure that most (though
	not quite all; see question 5.2) pointer arguments are converted
	correctly when passed as function arguments, so the question is
	largely moot.

	References: Rationale Sec. 4.1.5.

5.9:	If NULL and 0 are equivalent as null pointer constants, which
	should I use?

A:	Many programmers believe that NULL should be used in all pointer
	contexts, as a reminder that the value is to be thought of as a
	pointer.  Others feel that the confusion surrounding NULL and 0
	is only compounded by hiding 0 behind a macro, and prefer to use
	unadorned 0 instead.  There is no one right answer.  (See also
	questions 9.2 and 17.10.)  C programmers must understand that
	NULL and 0 are interchangeable in pointer contexts, and that an
	uncast 0 is perfectly acceptable.  Any usage of NULL (as opposed
	to 0) should be considered a gentle reminder that a pointer is
	involved; programmers should not depend on it (either for their
	own understanding or the compiler's) for distinguishing pointer
	0's from integer 0's.

	NULL should *not* be used when another kind of 0 is required,
	even though it might work, because doing so sends the wrong
	stylistic message.  (Furthermore, ANSI allows the definition of
	NULL to be ((void *)0), which will not work at all in non-
	pointer contexts.)  In particular, do not use NULL when the
	ASCII null character (NUL) is desired.  Provide your own
	definition

		#define NUL '\0'

	if you must.

	References: K&R1 Sec. 5.4 pp. 97-8; K&R2 Sec. 5.4 p. 102.

5.10:	But wouldn't it be better to use NULL (rather than 0), in case
	the value of NULL changes, perhaps on a machine with nonzero
	internal null pointers?

A:	No.  (Using NULL may be preferable, but not for this reason.)
	Although symbolic constants are often used in place of numbers
	because the numbers might change, this is *not* the reason that
	NULL is used in place of 0.  Once again, the language guarantees
	that source-code 0's (in pointer contexts) generate null
	pointers.  NULL is used only as a stylistic convention.  See
	questions 5.5 and 9.2.

5.12:	I use the preprocessor macro

		#define Nullptr(type) (type *)0

	to help me build null pointers of the correct type.

A:	This trick, though popular and superficially attractive, does
	not buy much.  It is not needed in assignments or comparisons;
	see question 5.2.  (It does not even save keystrokes.)  See also
	questions 9.1 and 10.2.

5.13:	This is strange.  NULL is guaranteed to be 0, but the null
	pointer is not?

A:	When the term "null" or "NULL" is casually used, one of several
	things may be meant:

	1.	The conceptual null pointer, the abstract language concept
		defined in question 5.1.  It is implemented with...

	2.	The internal (or run-time) representation of a null
		pointer, which may or may not be all-bits-0 and which may
		be different for different pointer types.  The actual
		values should be of concern only to compiler writers.
		Authors of C programs never see them, since they use...

	3.	The null pointer constant, which is a constant integer 0
		(see question 5.2).  It is often hidden behind...

	4.	The NULL macro, which is #defined to be 0 (see question
		5.4).  Finally, as red herrings, we have...

	5.	The ASCII null character (NUL), which does have all bits
		zero, but has no necessary relation to the null pointer
		except in name; and...

	6.	The "null string," which is another name for the empty
		string ("").  Using the term "null string" can be
		confusing in C, because an empty string involves a null
		('\0') character, but *not* a null pointer, which brings
		us full circle...

	This article uses the phrase "null pointer" (in lower case) for
	sense 1, the token "0" or the phrase "null pointer constant"
	for sense 3, and the capitalized word "NULL" for sense 4.

5.14:	Why is there so much confusion surrounding null pointers?  Why
	do these questions come up so often?

A:	C programmers traditionally like to know a lot (perhaps more
	than they need to) about the underlying machine implementation.
	The fact that null pointers are represented both in source code,
	and internally to most machines, as zero invites unwarranted
	assumptions.  The use of a preprocessor macro (NULL) may seem to
	suggest that the value could change some day, or on some weird
	machine.  The construct "if(p == 0)" is easily misread as
	calling for conversion of p to an integral type, rather than
	0 to a pointer type, before the comparison.  Finally, the
	distinction between the several uses of the term "null"
	(listed in question 5.13 above) is often overlooked.

	One good way to wade out of the confusion is to imagine that C
	used a keyword (perhaps "nil", like Pascal) as a null pointer
	constant.  The compiler could either turn "nil" into the
	appropriate type of null pointer when it could unambiguously
	determine that type from the source code, or complain when it
	could not.  Now in fact, in C the keyword for a null pointer
	constant is not "nil" but "0", which works almost as well,
	except that an uncast "0" in a non-pointer context generates an
	integer zero instead of an error message, and if that uncast 0
	was supposed to be a null pointer constant, the resulting
	program may not work.

5.15:	I'm confused.  I just can't understand all this null pointer
	stuff.

A:	Here are two simple rules you can follow:

	1.	When you want a null pointer constant in source code,
		use "0" or "NULL".

	2.	If the usage of "0" or "NULL" is an argument in a
		function call, cast it to the pointer type expected by
		the function being called.

	The rest of the discussion has to do with other people's
	misunderstandings, with the internal representation of null
	pointers (which you shouldn't need to know), and with the
	complexities of function prototypes.  (Taking those complexities
	into account, we find that rule 2 is conservative, of course;
	but it doesn't hurt.)  Understand questions 5.1, 5.2, and 5.4,
	and consider 5.3, 5.9, 5.13, and 5.14, and you'll do fine.

5.16:	Given all the confusion surrounding null pointers, wouldn't it
	be easier simply to require them to be represented internally by
	zeroes?

A:	If for no other reason, doing so would be ill-advised because it
	would unnecessarily constrain implementations which would
	otherwise naturally represent null pointers by special, nonzero
	bit patterns, particularly when those values would trigger
	automatic hardware traps for invalid accesses.

	Besides, what would such a requirement really accomplish?
	Proper understanding of null pointers does not require knowledge
	of the internal representation, whether zero or nonzero.
	Assuming that null pointers are internally zero does not make
	any code easier to write (except for a certain ill-advised usage
	of calloc(); see question 7.31).  Known-zero internal pointers
	would not obviate casts in function calls, because the *size* of
	the pointer might still be different from that of an int.  (If
	"nil" were used to request null pointers, as mentioned in
	question 5.14 above, the urge to assume an internal zero
	representation would not even arise.)

5.17:	Seriously, have any actual machines really used nonzero null
	pointers, or different representations for pointers to different
	types?

A:	The Prime 50 series used segment 07777, offset 0 for the null
	pointer, at least for PL/I.  Later models used segment 0, offset
	0 for null pointers in C, necessitating new instructions such as
	TCNP (Test C Null Pointer), evidently as a sop to all the extant
	poorly-written C code which made incorrect assumptions.  Older,
	word-addressed Prime machines were also notorious for requiring
	larger byte pointers (char *'s) than word pointers (int *'s).

	The Eclipse MV series from Data General has three
	architecturally supported pointer formats (word, byte, and bit
	pointers), two of which are used by C compilers: byte pointers
	for char * and void *, and word pointers for everything else.

	Some Honeywell-Bull mainframes use the bit pattern 06000 for
	(internal) null pointers.

	The CDC Cyber 180 Series has 48-bit pointers consisting of a
	ring, segment, and offset.  Most users (in ring 11) have null
	pointers of 0xB00000000000.  It was common on old CDC ones-
	complement machines to use an all-one-bits word as a special
	flag for all kinds of data, including invalid addresses.

	The old HP 3000 series uses a different addressing scheme for
	byte addresses than for word addresses; like several of the
	machines above it therefore uses different representations for
	char * and void * pointers than for other pointers.

	The Symbolics Lisp Machine, a tagged architecture, does not even
	have conventional numeric pointers; it uses the pair <NIL, 0>
	(basically a nonexistent <object, offset> handle) as a C null
	pointer.

	Depending on the "memory model" in use, 8086-family processors
	(PC compatibles) may use 16-bit data pointers and 32-bit
	function pointers, or vice versa.

	Some 64-bit Cray machines represent int * in the lower 48 bits
	of a word; char * additionally uses some of the upper 16 bits to
	indicate a byte address within a word.

	References: K&R1 Sec. A14.4 p. 211.

5.20:	What does a run-time "null pointer assignment" error mean?

A:	This message, which typically occurs with MS-DOS compilers,
	means that you've written, via a null pointer, to an invalid
	location -- probably offset 0 in the default data segment.
	See also question 16.8.


Section 6. Arrays and Pointers

6.1:	I had the definition char a[6] in one source file, and in
	another I declared extern char *a.  Why didn't it work?

A:	In one source file you defined an array of characters and in the
	other you declared a pointer to characters.  The declaration
	extern char *a simply does not match the actual definition.
	The type pointer-to-type-T is not the same as array-of-type-T.
	Use extern char a[].

	References: ISO Sec. 6.5.4.2; CT&P Sec. 3.3 pp. 33-4, Sec. 4.5
	pp. 64-5.

6.2:	But I heard that char a[] was identical to char *a.

A:	Not at all.  (What you heard has to do with formal parameters to
	functions; see question 6.4.)  Arrays are not pointers.  The
	array declaration char a[6] requests that space for six
	characters be set aside, to be known by the name "a".  That is,
	there is a location named "a" at which six characters can sit.
	The pointer declaration char *p, on the other hand, requests a
	place which holds a pointer, to be known by the name "p".  This
	pointer can point almost anywhere: to any char, or to any
	contiguous array of chars, or nowhere (see also questions 5.1
	and 1.30).

	As usual, a picture is worth a thousand words.  The declarations

		char a[] = "hello";
		char *p = "world";

	would initialize data structures which could be represented like
	this:
		   +---+---+---+---+---+---+
		a: | h | e | l | l | o |\0 |
		   +---+---+---+---+---+---+
		   +-----+     +---+---+---+---+---+---+
		p: |  *======> | w | o | r | l | d |\0 |
		   +-----+     +---+---+---+---+---+---+

	It is useful to realize that a reference like x[3] generates
	different code depending on whether x is an array or a pointer.
	Given the declarations above, when the compiler sees the
	expression a[3], it emits code to start at the location "a",
	move three past it, and fetch the character there.  When it sees
	the expression p[3], it emits code to start at the location "p",
	fetch the pointer value there, add three to the pointer, and
	finally fetch the character pointed to.  In other words, a[3] is
	three places past (the start of) the object *named* a, while
	p[3] is three places past the object *pointed to* by p.  In the
	example above, both a[3] and p[3] happen to be the character
	'l', but the compiler gets there differently.  (The essential
	difference is that the values of an array like a and a pointer
	like p are computed differently *whenever* they appear in
	expressions, whether or not they are being subscripted, as
	explained further in the next question.)  See also question 1.32.

	References: K&R2 Sec. 5.5 p. 104; CT&P Sec. 4.5 pp. 64-5.

6.3:	So what is meant by the "equivalence of pointers and arrays" in
	C?

A:	Much of the confusion surrounding arrays and pointers in C can
	be traced to a misunderstanding of this statement.  Saying that
	arrays and pointers are "equivalent" means neither that they are
	identical nor even interchangeable.  What it means is that array
	and pointer arithmetic is defined such that a pointer can be
	conveniently used to access an array or to simulate an array.

	Specifically, the cornerstone of the equivalence is this key
	definition:

		An lvalue of type array-of-T which appears in an
		expression decays (with three exceptions) into a
		pointer to its first element; the type of the
		resultant pointer is pointer-to-T.

	That is, whenever an array appears in an expression,
	the compiler implicitly generates a pointer to the array's
	first element, just as if the programmer had written &a[0].
	(The exceptions are when the array is the operand of a sizeof or
	& operator, or is a string literal initializer for a character
	array.)

	As a consequence of this definition, the compiler doesn't apply
	the array subscripting operator [] that differently to arrays
	and pointers, after all.  In an expression of the form a[i], the
	array decays into a pointer, following the rule above, and is
	then subscripted just as would be a pointer variable in the
	expression p[i] (although the eventual memory accesses will be
	different, as explained in question 6.2).  If you were to assign
	the array's address to the pointer:

		p = a;

	then p[3] and a[3] would access the same element.

	See also questions 6.8 and 6.14.

	References: K&R1 Sec. 5.3 pp. 93-6; K&R2 Sec. 5.3 p. 99; ISO
	Sec. 6.2.2.1, Sec. 6.3.2.1, Sec. 6.3.6; H&S Sec. 5.4.1 p. 124.

6.4:	Then why are array and pointer declarations interchangeable as
	function formal parameters?

A:	It's supposed to be a convenience.

	Since arrays decay immediately into pointers, an array is never
	actually passed to a function.  Allowing pointer parameters to
	be declared as arrays is a simply a way of making it look as
	though an array was being passed, perhaps because the parameter
	will be used within the function as if it were an array.
	Specifically, any parameter declarations which "look like"
	arrays, e.g.

		void f(char a[])
		{ ... }

	are treated by the compiler as if they were pointers, since that
	is what the function will receive if an array is passed:

		void f(char *a)
		{ ... }

	This conversion holds only within function formal parameter
	declarations, nowhere else.  If the conversion bothers you,
	avoid it; many programmers have concluded that the confusion it
	causes outweighs the small advantage of having the declaration
	"look like" the call or the uses within the function.

	See also question 6.21.

	References: K&R1 Sec. 5.3 p. 95, Sec. A10.1 p. 205; K&R2
	Sec. 5.3 p. 100, Sec. A8.6.3 p. 218, Sec. A10.1 

  


  

    
    
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