NASM Printing Out Time - Beginner Help

Question

I'm just getting started with ASM (NASM), and need some help with the following snippet. I get no errors/warnings, it just doesn't output anything. What I was expecting was for it to get the time (13), then print that out (4), then exit (1). Also, does anyone know of some good (preferably NASM specific) ASM tutorials?

section .bss
  time:   resb   255

section .text
  global _start

_start:
  mov   eax, 13
  int   0x80
  mov   eax, time

  mov   edx, 255
  mov   ecx, time
  mov   ebx, 1
  mov   eax, 4
  int   0x80

  mov   eax, 1
  int   0x80

I know this is probably terrible, but I'm just getting started. :)

Answer 1

This is your example translated to C. You are copying the pointer to time to eax instead of eax to the buffer. Still that wouldn't work because you want a char array for write and not a raw integer which will print garbage.

#include <stdlib.h>

char b[255];

int
main()
{
        /* You wanted to do this which doesn't work
         * because write wont take int* but char arrays
         * *(int*)b=time(NULL);
         */

        /* Instead you did */
        time(NULL);
        b;
        write(1, b, 255);
        exit(1);
}

Answer 2

Ok the best assembly book/tutorial at least for my is "http://drpaulcarter.com/pcasm", is a short tutorial focused in how the assembly works and how can be used with C/C++, explaining how this languages works. Is for x86 on 32 bits...