Re: no conversion from char[] to char* on function calls under circumstances [was: A bug?]

[Jan Engelhardt <jengelh at medozas dot de>]

On Tuesday 2008-12-16 17:05, Michel Van den Bergh wrote:

> Hi,
>
> The following program segfaults when compiled with gcc
> but runs fine when compiled with g++ or icc (the intel C compiler)
>
> #include <stdio.h>
> struct Hello {
>       char world[20];
> };
> struct Hello s(){
>       struct Hello r;
>       r.world[0]='H';
>       r.world[1]='\0';
>       return r;
> }
>
> int main(){
>       printf("%s\n",s().world);
> }
>
> Assigning s() to a variable and then using the variable avoids the segfault.

Had you compiled with -Wall would you have noticed:

	e.c:13: warning: format â%sâ expects type âchar *â, but
	argument 2 has type âchar[20]â

And when there is a type mismatch, a crash is pretty likely.
Not that I can say why gcc does not convert it to char* but g++ does.

Now what happens? The following augmented snippet shows it:
---<8---
#include <stdarg.h>
#include <stdio.h>
#include <string.h>
struct Hello {
       char world[20];
};
struct Hello s(void)
{
	struct Hello r;
	strcpy(r.world, "Hello");
	return r;
}
static void dump(const char *fmt, ...)
{
	va_list argp;
	va_start(argp, fmt);
	char *p = va_arg(argp, char *);
	printf("%p\n", p);
	va_end(argp);
}
int main(void)
{
	dump("", s().world);
	return 0;
}
--->8---

I get 0x6c6c6548, which is obviously part of the string Hello. So
passing a char[20] into a varargs function seems not to convert it to
char* when done through a non-visibile temporary (the result of s()
is hidden on the stack of main).