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From: tom@rlgvax.UUCP
Newsgroups: net.lang.c
Subject: Re: mixing pointers and arrays
Message-ID: <919@rlgvax.UUCP>
Date: Fri, 29-Jul-83 17:48:09 EDT
Article-I.D.: rlgvax.919
Posted: Fri Jul 29 17:48:09 1983
Date-Received: Mon, 1-Aug-83 10:46:27 EDT
References: <2332@csu-cs.UUCP>
Organization: CCI Office Systems Group, Reston, VA
Lines: 49

Concerning the following:

extern char *x;
main() {
	foo();
	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(foo);
	}
	subr(bar) char *bar; {
	}

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

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