理解 ARM 中的 str [英] Understanding str in ARM

问题描述

我有以下汇编代码:

alpha: .space 64    @reserves 64 bytes
i: int 0
   .text
   .align2
   .globalmain
   .func main
main:
      ldr r3,=i      @r3 is address of i
      mov r2,#0      @r2 is 0
      str r2,[r3]    @r2 is stored in r3
L1:
    ldr r3,=i             @r2 is address of i
    ldr r2,[r3]           @r2 is now address of i 
    cmp r2,#16            
    bge exit              @if r2 > 16, then exit (but it is not)
    ldr r3,=i             @r3 is address of i
    ldr r2,[r3]           @r2 is address of i
    add r1,r2,#0x200      @r1 is r2 + 0x200
    ldr r3,=alpha         @r3 is address of first element in array alpha (I think)
    str r1,[r3,r2,asl#2]
    ...

我知道很多，但我只有一部分有问题.我评论了我知道的部分，或者至少我认为我知道.

I know its alot, but I only have trouble with one part. I commented the parts I do know, or at least I think I know.

我遇到的问题是 str r1,[r3,r2,asl#2].我知道 asl#2 是左移，但这就是我真正知道的.r1 在 r3 中存储在哪里?str 命令的结果是什么?有人可以为我解释一下吗?

The trouble I'm having is with str r1,[r3,r2,asl#2]. I know asl#2 is a left shift, but that's all I really know. Where does r1 get stored, in r3? What would be the result of the str command? Can someone explain it for me?

推荐答案

str r1,[r3,r2,asl#2]

r2 中的值是算术左移(立即数)2 并添加到 r3 的内容中，以获得将存储 r1 的内存地址.

value in r2 is arithematic left shifted by (immediate value) 2 and is added to the contents of r3 , to get the memory address at which r1 will be stored.

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