I try to learn amd64 assembler. This is the first thing I tried. This piece of assembly should replicate the functionality of the following piece of C code, which turns a binary sha-256 hash into a human readable form.

assembly code

1:  .ascii "0123456789abcdef"
    .global show_hash
    .type show_hash @function
    .func show_hash
    mov $32,%ecx
    .p2align 2
0:  xor %eax,%eax
    mov %eax,%edx
    shr $4,%al
    and $15,%dl
    mov 1b(%rax),%al
    mov 1b(%rdx),%dl
    mov %bl,%ah
    dec %ecx
    jnz 0b
    mov %cl,(%rdi)

C code

void show_hash(char *dst, unsigned char *src) {
  static const char *lookup = "0123456789abcdef";
  char lo, hi, byte;
  int i = 32;

  do {
    byte = *src++;
    hi = lookup[byte >> 4];
    lo = lookup[byte & 0xf];
    *dst++ = hi;
    *dst++ = lo;
  } while (i--);

Am I doing it right? I tried to move the lookup table (label 1) into .section rodata, but all references to it were changed to 0 in the linked program, so I put it into the text section for now.

  • \$\begingroup\$ I don't think you can access data unless it's in the data segment. \$\endgroup\$
    – Hawken
    Oct 22, 2012 at 23:42
  • \$\begingroup\$ @Hawken How do you come to this conclusion? \$\endgroup\$
    – FUZxxl
    Oct 23, 2012 at 5:02
  • \$\begingroup\$ all memory is referenced using the segment specified by %ds or if it's on the stack, %ss. Unless you have personally set the segment register, mov %cs,%ds, or your assembler is adding a custom segment offset, you can't read from outside .data. I don't know if modern OSs allow you to read from .text at all even if you did. \$\endgroup\$
    – Hawken
    Oct 26, 2012 at 20:43
  • \$\begingroup\$ @Hawken You see that this code is written for x86-64 linux which uses a flat memory modell? All segment registers except %fs and %gs which serve a special purpose are set to zero. \$\endgroup\$
    – FUZxxl
    Oct 26, 2012 at 20:48
  • \$\begingroup\$ Hmm, I did not know that. wiki.osdev.org/X86-64 contains information on this for anyone else who did not know. \$\endgroup\$
    – Hawken
    Oct 26, 2012 at 21:43

1 Answer 1


I'm not sure whether it's correct:

  • If show_hash uses cdecl calling conventions you should preserve registers like %bl, and read the input parameter values from the stack into esi and edi.
    (see section "3.2.3 Parameter Passing" of the Application Binary Interface: parameters are conveniently passed in the rdi and rsi registers)
  • I don't understand the syntax of mov 1b(%rax),%al: is it reading from the lookup array defined at label 1:? On re-reading, I think it is; however that will only work if the most-significant bytes of rax and rdx are all zero; perhaps you should initialize them as you did using xor %eax,%eax
    (apparently xor %eax,%eax will clear the whole of %rax)
  • You're using stosw which writes two bytes (two ASCII characters) a time; but %ah contains a value from %bl, and %bl wasn't previously initialized? I think that statement should have been mov %dl,%ah not mov %bl,%ah; or perhaps you could have done mov 1b(%rdx),%ah directly.
  • The function ends with mov %cl,(%rdi) to null-terminate the string; that is clever but could use a comment (it took me a bit to figure out). Most of the questions either have good comments, or answer which say that they should have good comments.
  • \$\begingroup\$ xor %eax,%eax implicitly zeroes out the upper 32 bits of %rax as well. Second, this is x86_64 assembly on Linux. The calling convention is different; the first couple of arguments are actually passed in registers. I think I accidentially switched %dl with %bl, but I'm not very sury. \$\endgroup\$
    – FUZxxl
    Mar 6, 2014 at 20:12
  • \$\begingroup\$ I asked this question almost a year ago; I'm not really sure about most parts anymore. \$\endgroup\$
    – FUZxxl
    Mar 6, 2014 at 20:15
  • \$\begingroup\$ In that case the only error I see is the typo involving %bl instead of %dl. \$\endgroup\$
    – ChrisW
    Mar 6, 2014 at 20:40

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