数据移动错误说明 [英] data movement error clarification

问题描述

我目前正在解决第三版计算机系统中的问题3.3:程序员的观点，我很难理解这些错误的含义...

I'm currently solving problem 3.3 from 3rd edition of Computer System: a programmer's perspective and I'm having a hard time understanding what these errors mean...

movb $0xF, (%ebx)给出错误，因为ebx不能用作地址寄存器

movb $0xF, (%ebx) gives an error because ebx can't be used as address register

movl %rax, (%rsp)和 movb %si, 8(%rbp)给出错误，指出指令后缀和寄存器I.D之间不匹配.

movl %rax, (%rsp) and movb %si, 8(%rbp) gives error saying that theres a mismatch between instruction suffix and register I.D.

movl %eax, %rdx给出错误消息，表明目标操作数的大小不正确

movl %eax, %rdx gives an error saying that destination operand incorrect size

为什么我们不能使用ebx作为地址寄存器?是因为它的32位寄存器吗?如果不是movb $0xF, (%rbx)，下面的行是否可以工作?因为rbx是64位寄存器?

why can't we use ebx as address register? Is it because its 32-bit register? Would the following line work if it was movb $0xF, (%rbx) instead? since rbx is of 64bit register?

关于指令后缀和寄存器I.D之间不匹配的错误，是否出现此错误，因为它应该是movq %rax, (%rsp)和movew %si, 8(%rbp)而不是movl %rax, (%rsp)和movb %si, 8(%rbp)?

for the error regarding mismatch between instruction suffix and register I.D, does this error appear because it should've been movq %rax, (%rsp)and movew %si, 8(%rbp) instead of movl %rax, (%rsp) and movb %si, 8(%rbp)?

最后，对于与目标操作数大小不正确"有关的错误，这是因为目标寄存器是64位而不是32位吗?因此，如果代码行是movl %eax, %edx，则不会发生该错误?

and lastly, for the error regarding "destination operand incorrect size", is this because the destination register was 64 bit instead of 32? so if the line of code was movl %eax, %edx instead, the error wouldn't have occurred?

任何启发都会受到赞赏.

any enlightenment would be appreciated.

这是针对x86-64的

this is for x86-64

推荐答案

movb $0xF, (%ebx) gives an error because ebx can't be used as address register

确实，ebx不能用作地址寄存器(对于x86-64)，但是rbx可以. ebx是rbx的低32位. 64位代码的全部要点是地址可以是64位，因此尝试使用32位寄存器来引用内存毫无意义.

It's true that ebx can't be used as an address register (for x86-64), but rbx can. ebx is the lower 32bits of rbx. The whole point of 64bit code is that addresses can be 64bits, so trying to reference memory by using a 32bit register makes little sense.

movl %rax, (%rsp) and movb %si, 8(%rbp) gives error saying that 
theres a mismatch between instruction suffix and register I.D.

是的，因为您使用的是movl，所以'l'表示long，在这种情况下，它表示32位.但是，rax是一个64位寄存器.如果要从rax写入64位，则应使用movq.如果要写入32位，则应使用eax.

Yes, because you are using movl, the 'l' means long, which (in this context) means 32bits. However, rax is a 64bit register. If you want to write 64bits out of rax, you should use movq. If you want to write 32bits, you should use eax.

movl %eax, %rdx gives an error saying that destination operand incorrect size

您正在尝试将32位值移动到64位寄存器中.有说明可以为您执行此转换(例如，参见cdq)，但是movl并不是其中之一.

You are trying to move a 32bit value into a 64bit register. There are instructions to do this conversion for you (see cdq for example), but movl isn't one of them.

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