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

Question

What exactly does this instruction do?

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

Answer 1

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