What does the MOVZBL instruction do in IA-32 AT&T

问题:

What exactly the instruction

movzbl  0x01(%eax,%ecx),%eax

does?

回答1:

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.

回答2:

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

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

Runnable Intel example with assertions on GitHub.