mixing pointers and arrays

alan at allegra.UUCP alan at allegra.UUCP
Mon Aug 1 00:25:58 AEST 1983

Consider the following fragments:

	--- file 1 ---

	extern char *foo;

	func() {
		printf("%c", *foo);

	--- file 2 ---

	char foo[SIZE];


The declaration

	extern char *foo;

says that foo is a variable holding the address of a character (in
this case, the first character of an array).   This is not correct.
The variable foo doesn't refer to a pointer to an array, but to an
array itself.

On the other hand,

	main() {
		char foo[SIZE];


		char *foo;
		printf("%c", *foo);

will work just fine, although it would have been clearer to declare
foo as

	char foo[];

in func().

The key here is that in the first example, the foo in file one is the
same variable as the foo in file two, while in the second example, we
have two different foo's.   By saying


we create a character pointer.   In func(), foo now refers to a location
on the stack which holds the address of the first character of the array
known by the name foo in main().

Like so many other blunders, this is all done in the name of efficiency.
As far as I'm concerned, it's one of the ugliest parts of the C language.

						Alan Driscoll
						BTL, Murray Hill

More information about the Comp.lang.c mailing list