C++98 is not C99 :) there is no rvalue to lvalue conversion for rvalue
arrays in C++98. Also this code is still undefined C99 but will most
likely become valid C1x.
Sent from my iPhone
On Dec 16, 2008, at 8:45 AM, Jan Engelhardt <jeng...@medozas.de> wrote:
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 ‘%s’ expects type ‘char *’, but
argument 2 has type ‘char[20]’
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).
