了解在海峡ARM

问题描述

我有以下组装code：

 阿尔法：64。空间64 @reserves字节
我：INT 0
   。文本
   .align2
   .globalmain
   .FUNC主
主要：
      LDR R3，= I @ R3是我的地址
      MOV R2，＃0 @ r2为0
      海峡R2，[R3] // @ R2存在R3
L1：
    LDR R3，= I @ R2是我的地址
    LDR R2，[R3] // @ R2现在是我的地址
    CMP R2，＃16
    BGE退出@if R2＆GT; 16，然后退出（但不是）
    LDR R3，= I @ R3是我的地址
    LDR R2，[R3] // @ R2是我的地址
    添加R1，R2，＃为0x200 @ r1为R2 +为0x200
    LDR R3 =阿尔法@ R3是阿尔法数组第一个元素的地址（我认为）
    STR R1，[R3，R2，翔升＃2]
    ...
 

我知道它的很多，但我只有麻烦的一部分。我评论我知道的地方，或者至少我觉得我知道了。

我有麻烦的是 STR R1，[R3，R2，翔升＃2] 。我知道翔升＃2 是一个左移，但这是我真正知道了。哪里R1获取存储在R3？会是什么 STR 命令的结果？有人能解释一下吗？

解决方案

  STR R1，[R3，R2，翔升＃2]
 

在的r2值是由（立即值）移位arithematic左2，并加入到r3的内容，以获得在其中R1将被存储在存储器地址

I have the following assembly code:

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]
    ...

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.

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?

解决方案

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

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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