x86：将操作注册为内存内容和内存地址？ [英] x86: register operation as memory content and memory address?

问题描述

注册--->内存地址->内存内容

register ---> memory address --> memory content

内存地址->内存内容

以上模型正确吗？

，如果可以，您是否可以建议我是否认为正确？

and, if it is can you suggest if i am thinking it right or not??

movl％eax，％ebx->它将eax的内存地址移动到ebx，这也会导致内容也移动？

movl %eax, %ebx --> it moves memory address of eax to ebx which results in content moving as well??

movl（ ％eax），％ebx->它只是移动内容而不触摸地址??

movl (%eax),%ebx --> it just moves content and doesnot touch address??

移动值，％eax->将值的内存地址移动到％eax （以及内容，即eax具有新的内存内容）

movl value, %eax --> moves memory address of value to %eax (and content as well i.e. eax has new memory content)

移动％eax，val ---> val也具有新的内存地址和内容（均为％eax ）

movl %eax, val ---> val has new memory address and content as well(both of %eax)

movl $ val，％eax ---> val是移动到％eax内容的内容（地址不变）

movl $val, %eax ---> val is a content which is moved to content of %eax(address doesnt change)

movl％eax，$ val ---> ?????

movl %eax, $val ---> ?????

movl（％eax），val ---> val刚有新的内存内容

movl (%eax), val ---> val has just the new memory content

mov（％eax），$ val --->错误？

mov (%eax), $val ---> error?

mov $ val，（％eax）--->错误？

mov $val, (%eax) ---> error?

推荐答案

movl％eax，％ebx ->移动％eax 的>内容到寄存器％ebx 中。 ％eax 包含什么都无关紧要。

movl %eax, %ebx --> moves the content of register %eax into register %ebx. It is irrelevant what %eax contains.

movl（％eax），％ebx ->将％eax 中包含的内存地址的内容移至寄存器％ ebx

movl (%eax),%ebx --> moves the content of the memory address contained in %eax into register %ebx

移动值，％eax ->移动的内容内存地址 value 进入寄存器％eax 。

移动％eax，val ->将寄存器％eax 的内容移动到内存地址 val

movl %eax, val --> moves content of register %eax to memory address val

moval $ val，％eax ->将常量 $ val 移至寄存器％eax

move％eax，$ val -> 不可能。您不能将某物移动到恒定值。

movl %eax, $val --> impossible. You cannot move something into a constant value.

movl（％eax），val -> 不可能。您必须使用寄存器作为中间变量：

movl (%eax), val --> impossible. You must use a register as intermediate:

movl (%eax),%eax
movl %eax,val    ; or use another intermediate register than %eax                    

mov（％eax），$ val -> i 不可能。您不能将某物移动到恒定值。

mov (%eax), $val --> impossible. You cannot move something into a constant value.

mov $ val，（％eax）-> move 将常量值 $ val 转换为％eax 中包含的内存地址

mov $val, (%eax) --> move constant value $val into memory address contained in %eax.

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