What exactly does this instruction do?
movzbl 0x01(%eax,%ecx), %eax
44
votes
2 Answers
48
votes
AT&T syntax splits the movzx Intel instruction mnemonic into different mnemonics for different source sizes (movzb vs. movzw). In Intel syntax, it's:
movzx eax, byte ptr [eax+ecx+1]
i.e. load a byte from memory at eax+ecx+1 and zero-extend to full register.
BTW, most GNU tools now have a switch or a config option to prefer Intel syntax. (Such as objdump -Mintel or gcc -S -masm=intel, although the latter affects the syntax used when compiling inline-asm). I would certainly recommend to look into it, if you don't do AT&T assembly for living. See also the x86 tag wiki for more docs and guides.
28
votes
Minimal example
mov $0x01234567, %eax
mov $1, %bl
movzbl %bl, %eax
/* %eax == 0000 0001 */
mov $0x01234567, %eax
mov $-1, %bl
movzbl %bl, %eax
/* %eax == 0000 00FF */
Runanble GitHub upstream with assertions.
The mnemonic is:
- MOV
- Zero extend
- Byte (8-bit)
- to Long (32-bit)
There are also versions for other sizes:
movzbw: Byte (8-bit) to Word (16-bit)movzwl: Word (16-bit) to Long (32-bit)
Like most GAS instructions, you can omit the last size character when dealing with registers:
movzb %bl, %eax
but I cannot understand why we can't omit the before last letter, e.g. the following fails:
movz %bl, %eax
Why not just deduce it from the size of the operands when they are registers as for mov and Intel syntax?
And if you use registers of the wrong size, it fails to compile e.g.:
movzb %ax, %eax
movzbl 1(%eax,%ecx), %eax(that's the formatting GCC's asm output uses: godbolt.org/z/E4r9dP). This might have been literal compiler output or disassembly someone copy/pasted, with that spacing. – Peter Cordes