80386 - How does the cpu choose between movzbw and movzbl

Question

The opcodes for both movzbw and movzbl are OF B6. I don't understand how they can be distinguished by observing the ModR/M byte. From the Intel 80386 Programmer's Reference Manual (1986):

MOVZX  ── Move with Zero-Extend

Opcode    Instruction      Clocks   Description
0F B6 /r  MOVZX r16,r/m8   3/6       Move byte to word with zero-extend
0F B6 /r  MOVZX r32,r/m8   3/6       Move byte to dword, zero-extend
0F B7 /r  MOVZX r32,r/m16  3/6       Move word to dword, zero-extend

How does the processor distinguish between and movzbw and movzbl?

Answer 1

Looks like it uses a prefix byte:

66 0f b6 c0             movzx  ax,al
0f b6 c0                movzx  eax,al
0f b7 c0                movzx  eax,ax

Edit: note, in 64-bit mode, the above is the same but there is another prefix:

48 0f b6 c0             movzx  rax,al
48 0f b7 c0             movzx  rax,ax

Note that there is no movzx rax, eax instruction.

(I'm an utter novice at this so I can't explain exactly why, I just throw code at the compiler and see if it's accepted).