First steps with amd64 assembly

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

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.

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 assembly questions either have good comments, or answer which say that they should have good comments.

StackExchange Code Review Q#11110, answer score: 3