将较小的值移动到寄存器中 [英] Moving a value of a lesser size into a register

问题描述

我已经存储了 8 的一字节值，我想把它移到 rax 寄存器中.我目前正在使用 movzx 来对字节进行零扩展:

I have stored a one-byte value of 8 and I'd like to move that into the rax register. I'm currently doing this with movzx to zero-extend the byte:

.globl main
main:
    push %rbp
    mov %rsp, %rbp
    movb $8, -1(%rbp)
    movzx -1(%rbp), %rax <-- here
    ...

movzx 指令如何知道"-1(%rbp) 处的值只有一个字节长?从这里说，如果我正确阅读它，它可以在 byte 和 word 上工作，但它怎么知道?例如，如果我在 -2(%rbp) 处添加了一个两字节的值，它怎么知道要获取这个两字节的值?是否有另一条指令可以让我在地址处抓取 one 或 two 或 four 字节值并将其插入 64 位寄存器?

How does the movzx instruction 'know' that the value at -1(%rbp) is only one byte long? From here is says, if I'm reading it properly, that it can work on both a byte and a word, but how would it know? For example, if I added a two-byte value at -2(%rbp) how would it know to grab the two-byte value? Is there another instruction where I can just grab a one or two or four byte value at an address and insert it into a 64 bit register?

我想另一种方法是先将寄存器清零，然后将其添加到 8 位(或多位)组件中，例如:

I suppose another way to do it would be to first zero-out the register and then add it to the 8-bit (or however many bits) component, such as:

mov $0, %rax
mov -1(%rbp), %al

有没有一种方式比另一种方式更受欢迎?

Is there one way that is more preferred than another way?

推荐答案

movzx 指令如何知道"-1(%rbp) 处的值只有一个字节长?

How does the movzx instruction 'know' that the value at -1(%rbp) is only one byte long?

有两个(甚至三个)指令:

There are two (or even three) instructions:

movzxb(-1(%rbp) 是一个字节长)和 movzxw (-1(%rbp) 是一个 16 位字长).

movzxb (-1(%rbp) is one byte long) and movzxw (-1(%rbp) is one 16-bit word long).

我的汇编程序将 movzx 解释为 movzxb；但是，您不应该依赖它！

My assembler interprets movzx as movzxb; however, you should not rely on that!

最好使用包含源代码大小的指令名称(movzxb 或 movzxw)，以确保汇编程序使用正确的指令.

Better use the instruction name including the source size (movzxb or movzxw) to ensure that the assembler uses the correct instruction.

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