How to access a char from asm in c ,using asm intel syntax?

Question

The fallowing program must print 3 , but 'buff' is not seen from asm.

#include <stdio.h>

char buff[] = "%d\n";

int main (void)
{
    asm("mov eax, 3");
    asm("mov esi,eax");
    asm("mov edi,buff");
    asm("mov eax, 0");
    asm("call printf");
   return 0;
}

I try to use the asm intel syntax . it compile with : gcc -masm=intel -o test2 test2.c

The line

asm("mov edi,buff");

Is wrong , how do i must to write the buff here? I tried [buff] but do not works.Thank you

**Update:

next program works , BUT is using AT&T syntax :

#include <stdio.h>

char Format[] = "%d\n";

int main (void)
{
   asm
   (
      // Make stack space for arguments to printf
      "movl $3, %eax\n"
      "movl %eax, %esi\n"
      "movl $Format, %edi\n"
      "movl $0, %eax\n"
      "call printf\n"

   );
   return 0;
}  //compile with gcc -o test2 test2.c

I try to do same using intel syntax , but I do not know how to access the global variable ,from asm corectly

Answer 1

As @Peter Cordes said to me i read at https://stackoverflow.com/tags/x86/info about OFFSET keyword , and now it works. I knew the problem is i put in registry a data ,and not a address ,and looks like OFFSET keyword was what I was searching for .

Wrorking one is :

#include <stdio.h>

char buff[] = "%d\n";

int main (void)
{
    asm("mov eax, 3");
    asm("mov esi,eax");
    asm("mov edi,OFFSET buff");
    asm("mov eax, 0");
    asm("call printf");
   return 0;
}