This section discusses a very common use of pointers: setting things up so that a function can modify values in its caller, or return values, via its arguments. Remember that, normally, C passes arguments by value, and that if a function modifies one of its arguments, it modifies only its local copy, not the value in the caller. (Normally, this is a good thing; having a function which inadvertently assigns to its arguments and hence inadvertently modifies a value in its caller can be a source of obscure bugs in languages which don't use call-by-value.) However, what happens if a function wants to modify a value in its caller, and its caller wants to let it? How can a function return two values? (A function's formal return value is always a single value.)
The answer to both questions is that a function can declare a parameter which is a pointer. The caller then passes a pointer to (that is, the ``address of'') a variable in the caller which is to be modified or which is to receive the second return value. In fact, we've seen examples of this already: getline returns the length of the line it reads as well as the line itself, and the getop routine in the calculator example in section 4.3 returned both a code for an operator and a string representing the full text of the operator. (We needed that string when the operator was '0' indicating numeric input, so that the string could return the full numeric input.) Though we didn't say so at the time, we were actually using pointers in these examples. (We'll explore the relationship between arrays and pointers, which made this possible, in section 5.3.)
With all of this in mind, make sure that you understand why the swap example on page 95 would not work, and how and why the swap example on page 96 does work, and what the figure on page 96 shows.
The swap example demonstrated a function which modified some variables (a and b) in its caller. The getint example demonstrates how to return two values from a function by returning one value as the normal function return value and the other one by writing to a pointer. (There is no fundamental difference, though, between ``modifying a variable in the caller'' and ``returning a value by writing to a pointer''; these are just two applications of pointer parameters.)
The version of getint on page 97 is somewhat complicated because it allows free-form input, that is, the values need only be separated by whitespace or punctuation; they are not restricted to being one per line or anything. (C source code is also free-form in this regard; see page 4 of chapter 1 of these notes.) To see more clearly the essence of what getint is supposed to do, imagine for a moment that the input is restricted to one value per line, as in the ``primitive calculator'' example on page 72 of section 4.2. In that case, we might use the following simpler (i.e. more primitive) code:
int getint(int *pn)
{
char line[20];
if (getline(line, 20) <= 0)
return EOF;
*pn = atoi(line);
return 1;
}
The getint function on page 97 is
documented as returning nonzero for a valid number and 0 for
something other than a number.
Our stripped-down version does not,
and as it happens,
the example code at the bottom of page 96 does not make use of
the valid/invalid distinction.
Can you see a way to rewrite the code at the bottom of page 96
to fill in the cells of the array with only valid numbers?
You might also notice, again from the code at the bottom of page 96, that & need not be restricted to single, simple variables; it can take the address of any data object, in this case, one cell of the array. Just as for all of C's other operators, & can be applied to arbitrary expressions, although it is restricted to expressions which represent addressable objects. Expressions like &1 or &(2+3) are meaningless and illegal.
You may remember a discussion from section 1.5.1 on page 16 of how C's getchar routine is able to return all possible characters, plus an end-of-file indication, in its single return value. Why does getint need two return values? Why can't it use the same trick that getchar does?
The examples in this section are again relatively safe. The pointers have all been parameters, and the callers have passed pointers (that is, the ``addresses'' of) their own, properly-allocated variables. That is, code like
int a = 1, b = 2; swap(&a, &b);and
int a; getint(&a);is correct and quite safe.
Something to beware of, though, is the temptation to inadvertently pass an uninitialized pointer variable (rather than the ``address'' of some other variable) to a routine which expects a pointer. We know that the getint routine expects as its argument a pointer to an int in which it is to store the integer it gets. Suppose we took that description literally, and wrote
int *ip; /* a pointer to an int */ getint(ip);Here we have in fact passed a pointer-to-int to getint, but the pointer we passed doesn't point anywhere! When getint writes to (``dereferences'') the pointer, in an expression like *pn = 0, it will scribble on some random part of memory, and the program may crash. When people get caught in this trap, they often think that to fix it they need to use the & operator:
getint(&ip); /* WRONG */or maybe the * operator:
getint(*ip); /* WRONG */but these go from bad to worse. (If you think about them carefully, &ip is a pointer-to-pointer-to-int, and *ip is an int, and neither of these types matches the pointer-to-int which getint expects.) The correct usage for now, as we showed already, is something like
int a; getint(&a);In this case, a is an honest-to-goodness, allocated int, so when we generate a pointer to it (with &a) and call getint, getint receives a pointer that does point somewhere.
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This page by Steve Summit // Copyright 1995, 1996 // mail feedback