错误:"movq"的操作数大小不匹配 [英] Error: operand size mismatch for `movq'

问题描述

我正在尝试编译以下程序集...

I'm trying to compile the following assembly...

movq $0x3534373536383235, 0x000000000055638f8
movq $0x55638f8, %rdi
retq

第一行引发错误Error: operand size mismatch for 'movq'

这对我来说没有意义，因为它们都是8字节数字.

Which doesn't make sense to me, because they are both 8 byte numbers.

我做了一些研究，并推荐了movabsq，就像这样...

I did a little research and movabsq was recommended, like so...

movabsq $0x3534373536383235, 0x000000000055638f8
movq $0x55638f8, %rdi
retq

但这会引发错误:Error: operand size mismatch for 'movabs'

我想念什么?

这是我的Mac上的全部错误

Here's the entire error from my mac

level3.s:1:27: error: invalid operand for instruction
movq $0x3534373536383235, 0x000000000055638f8
                          ^~~~~~~~~~~~~~~~~~~

推荐答案

如果您在引理MOV下查看该手册(诚然有点吓人)，您会发现没有mov r/m64, imm64

If you look in the manual under the lemma MOV (which is, admittedly, a bit intimidating), you will find that there is no mov r/m64, imm64.

有一种方法可以将完整的64位常量加载到寄存器中，也可以将符号扩展的32位常量加载到64位内存位置中，但是没有一条指令可以立即使用64位并将其写入内存

There's a way to load a full 64bit constant into a register, and there's a way to load a sign-extended 32bit constant into a 64bit memory location, but there is no single instruction that takes a 64bit immediate and writes it to memory.

因此，您必须分两个步骤进行操作，例如通过使用临时寄存器或直接将两个双字直接写入内存.

So you have to do it in two steps, such as by using a temporary register or by writing two dwords straight to memory separately.

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