NASM-如何将8位寄存器移到完整的32位寄存器中? [英] NASM - How do you move an 8-bit register into a full 32-bit register?

问题描述

我正在编写NASM汇编代码，并且必须执行一些索引寻址.我已经将索引存储在$ al中，但是x86不允许您使用$ al作为索引寄存器，并且我已经在使用$ bl，因此我不能使用$ bx.因此，我需要将$ al中的字节放入32个但可注册的寄存器中，例如$ ecx，但是，当我尝试时，它会抛出操作码和操作数错误的无效组合.有什么办法吗?

I am writing NASM assembly code, and have to do some indexed addressing. I have the index stored in $al, but x86 won't let you use $al as an index register, and I'm already using $bl, so I cant use $bx. So I need to put the byte I have in $al into a 32-but register such as $ecx, however, when I try, it throws an invalid combination of opcode and operand error. Is there any way to do this?

    sub     al, 97                  ; char - 97

    push    ecx                     ; b/c al cant be used as indexing register
    mov     ecx, al                 ; move byte in al into ecx

    mov     bl, [table + ecx]       ; value_at(first_table_addr + char) -> bx

    pop     ecx

推荐答案

使用 MOVZX指令:

movzx ecx, al  ; move byte to doubleword, zero-extension

如果您希望将al中的值视为带符号，则还有MOVSX.

There's also MOVSX if you want the value in al to be treated as signed.

零扩展表示目标操作数的高位将被设置为零，而符号扩展表示目标操作数的高位将被设置为源操作数的符号.一些例子:

Zero-extention means the upper bits of the destination operand will be set to zero, while sign-extension means the upper bits of the destination operand will be set to the sign bit of the source operand. Some examples:

mov al,0x7F
movzx ebx,al   ; ebx = 0x0000007F
movsx ebx,al   ; ebx = 0x0000007F

mov al,0x80
movzx ebx,al   ; ebx = 0x00000080
movsx ebx,al   ; ebx = 0xFFFFFF80

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