mixing pointers and arrays

tom at rlgvax.UUCP tom at rlgvax.UUCP
Sat Jul 30 07:48:09 AEST 1983

Concerning the following:

extern char *x;
main() {
	printf("%s", *x);
-----------------------------<new file>----------
char x[10];
sub() {
	x[0] = 'a';
	x[1] = '\0';

First, the reason the program won't work is that it should be
"printf("%s", x);" WITHOUT the indirection on x.  What was originally
written passes a single character to printf() where a pointer (address)
was expected.

Second, the declaration "extern char *x;" is incorrect.  "x" is an array,
NOT a pointer, and must be declared as such.  The compiler thinks that
x is a pointer (i.e. a storage location containing an address).  When
you pass x to printf(), x is evaulated, i.e. the contents of that location
is pushed onto the stack.  However, in reality x is an ARRAY!  This means
that what "the contents of that location" is in fact a bunch of characters.
So, you are passing a bunch of characters which are being interpeted as
an address.  So, you blow up.  Solution: in main, declare "extern char x[];".

The example you mention:

	char *foo, bar[10];
	foo = bar;

works because using "bar" by itself is exactly the same as "&(bar[0])"
by definition in C, not because you can treat pointers and arrays
indiscriminately.  This next example also works for the same reason.

	main() {
	char foo[10];
	subr(bar) char *bar; {

I find that most new programmers find this second example confusing to say the

- Tom Beres
{seismo, allegra, brl-bmd, mcnc, we13}!rlgvax!tom

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