Doubt about char in C

Hi,

Code:

---

#include<stdio.h>

struct an
{
 int num;
 char name[256];
}a;

char *match = "This is a test: 10 ok";

int main(void) {
sscanf(match, "This is a test: %d %s", &a.num, &a.name);
printf("%d\n%s\n", a.num, a.name);
return 0;
}

---

When I compile the above program I am getting the below warning.

Code:

---

t1.c: In function ‘main’:
t1.c:13: warning: format ‘%s’ expects type ‘char *’, but argument 4 has type ‘char (*)[256]’

---

why I am getting this warning, I tried removing the size of char name but printf throwing similar warning, what is causing this warning and how to clear this warning?


Thanks in advance,
Alagunambi Welkin

I am using gcc

The name of an array is used as a pointer to it's first element, so you shouldn't be using the & (address of) operator with it -- that will give you a pointer to a pointer.

Ie, just lose the & with a.name