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Frequently Asked Questions (FAQS);faqs.221



Besides listing frequently-asked questions, this article also summarizes
frequently-posted answers.  Even if you know all the answers, it's worth
skimming through this list once in a while, so that when you see one of
its questions unwittingly posted, you won't have to waste time
answering.

This article is always being improved.  Your input is welcomed.  Send
your comments to [e-mail], [e-mail], and/or
mit-eddie!adam.mit.edu!scs; this article's From: line may be unusable.

The questions answered here are divided into several categories:

     1. Null Pointers
     2. Arrays and Pointers
     3. Memory Allocation
     4. Expressions
     5. ANSI C
     6. C Preprocessor
     7. Variable-Length Argument Lists
     8. Boolean Expressions and Variables
     9. Structs, Enums, and Unions
    10. Declarations
    11. Stdio
    12. Library Subroutines
    13. Lint
    14. Style
    15. Floating Point
    16. System Dependencies
    17. Miscellaneous (Fortran to C converters, YACC grammars, etc.)

Herewith, some frequently-asked questions and their answers:


Section 1. Null Pointers

1.1:    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 not
    the address of any object.  That is, the address-of operator &
    will never yield a null pointer, nor will a successful call to
    malloc.  (malloc returns 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; an
    uninitialized pointer might point anywhere.  See also questions
    3.1, 3.9, and 17.1.

    As mentioned in the definition 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 it can make
    the distinction if necessary (see below).

    References: K&R I Sec. 5.4 pp. 97-8; K&R II Sec. 5.4 p. 102; H&S
    Sec. 5.3 p. 91; ANSI Sec. 3.2.2.3 p. 38.

1.2:    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)

    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.  For instance, the Unix system call "execl" takes a
    variable-length, null-pointer-terminated list of character
    pointer arguments.  To generate a null pointer in a function
    call context, an explicit cast is typically required, to force
    the 0 to be in a pointer context:

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

    If the (char *) cast 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
    unadorned 0's.  Function prototypes cannot provide the types for
    variable arguments in variable-length argument lists, however,
    so explicit casts are still required for those arguments.  It is
    safest always to cast null pointer function arguments, to guard
    against varargs functions or those without prototypes, to allow
    interim use of non-ANSI compilers, and to demonstrate that you
    know what you are doing.  (Incidentally, it's also a simpler
    rule to remember.)

    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&R I Sec. A7.7 p. 190, Sec. A7.14 p. 192; K&R II
    Sec. A7.10 p. 207, Sec. A7.17 p. 209; H&S Sec. 4.6.3 p. 72; ANSI
    Sec. 3.2.2.3 .

1.3:    What is NULL and how is it #defined?

A:    As a matter of style, many people prefer not to have unadorned
    0's scattered throughout their programs.  For this reason, the
    preprocessor macro NULL is #defined (by <stdio.h> or
    <stddef.h>), with value 0 (or (void *)0, about which more
    later).  A programmer who wishes to make explicit the
    distinction between 0 the integer and 0 the null pointer can
    then use NULL whenever a null pointer is required.  This is a
    stylistic convention only; the preprocessor turns NULL back to 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 1.2 above applies for NULL as well as 0.)

    NULL should _only_ be used for pointers; see question 1.8.

    References: K&R I Sec. 5.4 pp. 97-8; K&R II Sec. 5.4 p. 102; H&S
    Sec. 13.1 p. 283; ANSI Sec. 4.1.5 p. 99, Sec. 3.2.2.3 p. 38,
    Rationale Sec. 4.1.5 p. 74.

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

A:    Programmers should never need to know the internal
    representation(s) of null pointers, because they are normally
    taken care of by the compiler.  If a machine uses a nonzero bit
    pattern for null pointers, it is the compiler's responsibility
    to generate it when the programmer requests, by writing "0" or
    "NULL," a null pointer.  Therefore, #defining NULL as 0 on a
    machine for which internal null pointers are nonzero is as valid
    as on any other, because the compiler must (and can) still
    generate the machine's correct null pointers in response to
    unadorned 0's seen in pointer contexts.

1.5:    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 general.  The problem 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 to work
    correctly, but pointer arguments to other types would still be
    problematical, and legal constructions such as

        FILE *fp = NULL;

    could fail.

    Nevertheless, ANSI C allows the alternate

        #define NULL ((void *)0)

    definition for NULL.  Besides 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 1.8).

1.6:    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 in some circles, does not buy much.
    It is not needed in assignments and comparisons; see question
    1.2.  It does not even save keystrokes.  Its use suggests to the
    reader that the author is shaky on the subject of null pointers,
    and requires the reader to check the #definition of the macro,
    its invocations, and _all_ other pointer usages much more
    carefully.  See also question 8.1.

1.7:    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 (in the if,
    while, for, and do statements, and with the &&, ||, !, and ?:
    operators), a false value is produced 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 a null pointer, and use the correct value.
    There is no trickery involved here; compilers do work this way,
    and generate identical code for both statements.  The internal
    representation of a pointer does _not_ matter.

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

        !expr    is essentially equivalent to    expr?0:1

    It is left as an exercise for the reader to show 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.

    See also question 8.2.

    References: K&R II Sec. A7.4.7 p. 204; H&S Sec. 5.3 p. 91; ANSI
    Secs. 3.3.3.3, 3.3.9, 3.3.13, 3.3.14, 3.3.15, 3.6.4.1, and
    3.6.5 .

1.8:    If "NULL" and "0" are equivalent, 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
    #definition, and prefer to use unadorned "0" instead.  There is
    no one right answer.  C programmers must understand that "NULL"
    and "0" are interchangeable and that an uncast "0" is perfectly
    acceptable in initialization, assignment, and comparison
    contexts.  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.  (ANSI allows the #definition of NULL to be
    (void *)0, which will not work 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.

    Reference: K&R II Sec. 5.4 p. 102.

1.9:    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
    null pointers?

A:    No.  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.

1.10:    I'm confused.  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 1.1.  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 source code syntax for null pointers, which is the
        single character "0".  It is often hidden behind...

    4.    The NULL macro, which is #defined to be "0" or
        "(void *)0".  Finally, as red herrings, we have...

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

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

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

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

A:    C programmers traditionally like to know 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) suggests that the value might
    change later, 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 above) is often overlooked.

    One good way to wade out of the confusion is to imagine that C
    had a keyword (perhaps "nil", like Pascal) with which null
    pointers were requested.  The compiler could either turn "nil"
    into the correct type of null pointer, when it could determine
    the type from the source code, or complain when it could not.
    Now, in fact, in C the keyword for a null pointer 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, the code may not work.

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

A:    Follow these two simple rules:

    1.    When you want to refer to a null pointer 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, or with the internal representation of null
    pointers (which you shouldn't need to know), or with ANSI C
    refinements.  Understand questions 1.1, 1.2, and 1.3, and
    consider 1.8 and 1.11, and you'll do fine.

1.13:    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 this 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 3.9).  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 rather than "0," as mentioned in question
    1.11, the urge to assume an internal zero representation would
    not even arise.)

1.14:    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).

    Some Honeywell-Bull mainframes use the bit pattern 06000 for
    (internal) null 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, 80*86 processors (PC's)
    may use 16 bit data pointers and 32 bit function pointers, or
    vice versa.


Section 2. Arrays and Pointers

2.1:    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:    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: CT&P Sec. 3.3 pp. 33-4, Sec. 4.5 pp. 64-5.

2.2:    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 2.4.)  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.  The pointer is to be known by the
    name "p," and can point to any char (or contiguous array of
    chars) anywhere.

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

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

    would result in data structures which could be represented like
    this:

           +---+---+---+---+---+---+
        a: | h | e | l | l | o |\0 |
           +---+---+---+---+---+---+

           +-----+     +---+---+---+---+---+---+
        p: |  *======> | w | o | r | l | d |\0 |
           +-----+     +---+---+---+---+---+---+

    It is important to remember 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 the example above,
    both a[3] and p[3] happen to be the character 'l', but the
    compiler gets there differently.  (See also question 17.14.)

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

A:    Much of the confusion surrounding pointers in C can be traced to
    a misunderstanding of this statement.  Saying that arrays and
    pointers are "equivalent" does not by any means imply that they
    are interchangeable.

    "Equivalence" refers to the following 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.

    (The exceptions are when the array is the operand of a sizeof or
    & operator, or is a literal string initializer for a character
    array.)

    As a consequence of this definition, there is not really any
    difference in the behavior of the "array subscripting" operator
    [] as it applies to arrays and pointers.  In an expression of
    the form a[i], the array reference "a" decays into a pointer,
    following the rule above, and is then subscripted exactly as
    would be a pointer variable in the expression p[i].  In either
    case, the expression x[i] (where x is an array or a pointer) is,
    by definition, exactly equivalent to *((x)+(i)).

    References: K&R I Sec. 5.3 pp. 93-6; K&R II Sec. 5.3 p. 99; H&S
    Sec. 5.4.1 p. 93; ANSI Sec. 3.2.2.1, Sec. 3.3.2.1, Sec. 3.3.6 .

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

A:    Since arrays decay immediately into pointers, an array is never
    actually passed to a function.  As a convenience, any parameter
    declarations which "look like" arrays, e.g.

        f(a)
        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:

        f(a)
        char *a;

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

    References: K&R I Sec. 5.3 p. 95, Sec. A10.1 p. 205; K&R II
    Sec. 5.3 p. 100, Sec. A8.6.3 p. 218, Sec. A10.1 p. 226; H&S
    Sec. 5.4.3 p. 96; ANSI Sec. 3.5.4.3, Sec. 3.7.1, CT&P Sec. 3.3
    pp. 33-4.

2.5:    Why doesn't sizeof properly report the size of an array which is
    a parameter to a function?

A:    The sizeof operator reports the size of the pointer parameter
    which the function actually receives (see question 2.4).

2.6:    Someone explained to me that arrays were really just constant
    pointers.

A:    This is a bit of an oversimplification.  An array name is
    "constant" in that it cannot be assigned to, but an array is
    _not_ a pointer, as the discussion and pictures in question 2.2
    should make clear.

2.7:    I came across some "joke" code containing the "expression"
    5["abcdef"] .  How can this be legal C?

A:    Yes, Virginia, array subscripting is commutative in C.  This
    curious fact follows from the pointer definition of array
    subscripting, namely that a[e] is exactly equivalent to
    *((a)+(e)), for _any_ expression e and primary expression a, as
    long as one of them is a pointer expression and one is integral.
    This unsuspected commutativity is often mentioned in C texts as
    if it were something to be proud of, but it finds no useful
    application outside of the Obfuscated C Contest (see question
    17.9).

    References: ANSI Rationale Sec. 3.3.2.1 p. 41.

2.8:    My compiler complained when I passed a two-dimensional array to
    a routine expecting a pointer to a pointer.

A:    The rule by which arrays decay into pointers is not applied
    recursively.  An array of arrays (i.e. a two-dimensional array
    in C) decays into a pointer to an array, not a pointer to a
    pointer.  Pointers to arrays can be confusing, and must be
    treated carefully.  (The confusion is heightened by the
    existence of incorrect compilers, including some versions of pcc
    and pcc-derived lint's, which improperly accept assignments of
    multi-dimensional arrays to multi-level pointers.)  If you are
    passing a two-dimensional array to a function:

        int array[YSIZE][XSIZE];
        f(array);

    the function's declaration should match:

        f(int a[][XSIZE]) {...}
    or
        f(int (*ap)[XSIZE]) {...}    /* ap is a pointer to an array */

    In the first declaration, the compiler performs the usual
    implicit parameter rewriting of "array of array" to "pointer to
    array;" in the second form the pointer declaration is explicit.
    Since the called function does not allocate space for the array,
    it does not need to know the overall size, so the number of
    "rows," YSIZE, can be omitted.  The "shape" of the array is
    still important, so the "column" dimension XSIZE (and, for 3- or
    more dimensional arrays, the intervening ones) must be included.

    If a function is already declared as accepting a pointer to a
    pointer, it is probably incorrect to pass a two-dimensional
    array directly to it.

    References: K&R I Sec. 5.10 p. 110; K&R II Sec. 5.9 p. 113.

2.9:    How do I declare a pointer to an array?

A:    Usually, you don't want to.  Consider using a pointer to one of
    the array's elements instead.  Arrays of type T decay into
    pointers to type T (see question 2.3), which is convenient;
    subscripting or incrementing the resultant pointer accesses the
    individual members of the array.  True pointers to arrays, when
    subscripted or incremented, step over entire arrays, and are
    generally only useful when operating on arrays of arrays, if at
    all.  (See question 2.8 above.)  When people speak casually of a
    pointer to an array, they usually mean a pointer to its first
    element.

    If you really need to declare a pointer to an entire array, use
    something like "int (*ap)[N];" where N is the size of the array.
    (See also question 10.3.)  If the size of the array is unknown,
    N can be omitted, but the resulting type, "pointer to array of
    unknown size," is useless.

2.10:    How can I dynamically allocate a multidimensional array?

A:    It is usually best to allocate an array of pointers, and then
    initialize each pointer to a dynamically-allocated "row."  The
    resulting "ragged" array can save space, although it is not
    necessarily contiguous in memory as a real array would be.  Here
    is a two-dimensional example:

        int **array = (int **)malloc(nrows * sizeof(int *));
        for(i = 0; i < nrows; i++)
            array[i] = (int *)malloc(ncolumns * sizeof(int));

    (In "real" code, of course, malloc would be declared correctly,
    and each return value checked.)

    You can keep the array's contents contiguous, while making later
    reallocation of individual rows difficult, with a bit of
    explicit pointer arithmetic:

        int **array = (int **)malloc(nrows * sizeof(int *));
        array[0] = (int *)malloc(nrows * ncolumns * sizeof(int));
        for(i = 1; i < nrows; i++)
            array[i] = array[0] + i * ncolumns;

    In either case, the elements of the dynamic array can be
    accessed with normal-looking array subscripts: array[i][j].

    If the double indirection implied by the above schemes is for
    some reason unacceptable, you can simulate a two-dimensional
    array with a single, dynamically-allocated one-dimensional
    array:

        int *array = (int *)malloc(nrows * ncolumns * sizeof(int));

    However, you must now perform subscript calculations manually,
    accessing the i,jth element with array[i * ncolumns + j].  (A
    macro can hide the explicit calculation, but invoking it then
    requires parentheses and commas which don't look exactly like
    multidimensional array subscripts.)

    Finally, you can use pointers-to-arrays:

        int (*array)[NCOLUMNS] =
            (int (*)[NCOLUMNS])malloc(nrows * sizeof(*array));

    , but the syntax gets horrific and all but one dimension must be
    known at compile time.

    With all of these techniques, you may of course need to remember
    to free the arrays (which may take several steps) when they are
    no longer needed.  You must also be extremely cautious when
    passing dynamically-allocated arrays down to other functions, if
    those functions are also to accept conventional, statically-
    allocated arrays (see question 2.8).

2.11:    Here's a neat trick: if I write

        int realarray[10];
        int *array = &realarray[-1];

    I can treat "array" as if it were a 1-based array.

A:    This technique, though attractive, is not strictly conforming to
    the C standards (in spite of its appearance in Numerical Recipes
    in C).  Pointer arithmetic is defined only as long as the
    pointer points within the same allocated block of memory, or to
    the imaginary "terminating" element one past it; otherwise, the
    behavior is undefined, _even if the pointer is not
    dereferenced_.  The code above could fail if, while subtracting
    the offset, an illegal address were generated (perhaps because
    the address tried to "wrap around" past the beginning of some
    memory segment).

    References: ANSI Sec. 3.3.6 p. 48, Rationale Sec. 3.2.2.3 p. 38;
    K&R II Sec. 5.3 p. 100, Sec. 5.4 pp. 102-3, Sec. A7.7 pp. 205-6.

2.12:    I passed a pointer to a function which initialized it:

        main()
        {
            int *ip;
            f(ip);
            return 0;
        }

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

    , but the pointer in the caller was unchanged.

A:    Did the function try to initialize the pointer itself, or just
    what it pointed to?  Remember that arguments in C are passed by
    value.  The called function altered only the passed copy of the
    pointer.  You'll want to pass the address of the pointer (the
    function will end up accepting a pointer-to-a-pointer).

2.13:    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 an anomaly in pcc-
    derived compilers, and an extension in gcc, that expressions
    such as the above are ever accepted.)  Say what you mean: use