如何执行本节代码，但在ARM Assembly中使用自动索引 [英] How can I do this section of code, but using auto-indexing with ARM Assembly

问题描述

这行得通，但是我必须使用自动索引来做到这一点，我无法弄清楚那部分.

this works, but I have to do it using auto-indexing and I can not figure out that part.

writeloop:
  cmp r0, #10
  beq writedone
  ldr r1, =array1
  lsl r2, r0, #2
  add r2, r1, r2
  str r2, [r2]
  add r0, r0, #1
  b writeloop

还有我拥有的数据

.balign 4
array1: skip 40

我尝试过的是这个，是的，我知道这可能是一个糟糕的尝试，但是我对此并不陌生

What I had tried was this, and yes I know it is probably a poor attempt but I am new to this and do not understand

ldr r1, =array1

writeloop:
  cmp r0, #10
  beq writedone
  ldr r2, [r1], #4
  str r2, [r2]
  add r0, r0, #1
  b writeloop

尝试此操作时显示细分错误.怎么了?我想应该发生的是，每次循环通过时，都会将元素r2的=设置为自身的地址，然后递增到下一个元素并执行相同的操作

It says segmentation fault when I try this. What is wrong? What I am thinking should happen is every time it loops through, it sets the element r2 it at = to the address of itself, and then increments to the next element and does the same thing

推荐答案

ARM体系结构提供了几种不同的地址模式.

The ARM architechures gives several different address modes.

来自ARM946E-S产品概述和许多其他来源:

加载和存储指令具有三种主要寻址模式
-偏移量
-预索引
-后索引.

它们是通过在基址寄存器中添加或减去立即数或基于寄存器的偏移量而形成的.基于寄存器的偏移量也可以通过移位操作进行缩放.预索引和后索引寻址模式使用基加偏移量更新来更新基址寄存器.由于PC是通用寄存器，因此可以将32位值直接加载到PC中，以跳转到4GB内存空间中的任何地址.

Load and store instructions have three primary addressing modes
- offset
- pre-indexed
- post-indexed.

They are formed by adding or subtracting an immediate or register-based offset to or from a base register. Register-based offsets can also be scaled with shift operations. Pre-indexed and post-indexed addressing modes update the base register with the base plus offset calculation. As the PC is a general purpose register, a 32‑bit value can be loaded directly into the PC to perform a jump to any address in the 4GB memory space.

同样，它们支持回写或寄存器更新，因此需要预索引和后索引.如果没有回写，后索引没有多大意义.

As well, they support write back or updating of the register, hence the reason for pre-indexed and post-indexed. Post-index doesn't make much sense without write back.

由于您遇到的问题，我相信您希望将值0-9写入由10个字组成的数组(长度为4个字节).假设这样，您可以使用索引并通过 add 更新值.导致

Now to your issue, I believe that you want to write the values 0-9 to an array of ten words (length four bytes). Assuming this, you can use indexing and update the value via add. This leads to,

  mov r0, #0       ; start value
  ldr r1, =array1  ; array pointer

writeloop:
  cmp r0, #10
  beq writedone
  str r0, [r1, r0, lsl #2] ; index with r1 base by r0 scaled by *4
  add r0, r0, #1
  b writeloop
writedone:
; code to jump somewhere else and not execute data.

.balign 4
array1: skip 40

出于兴趣，可以通过计数和写下来完成更有效的循环

For interest a more efficient loop can be done by counting and writing down,

  mov r0, #9       ; start value
  ldr r1, =array1  ; array pointer

writeloop:
  str r0, [r1, r0, lsl #2] ; index with r1 base by r0 scaled by *4
  subs r0, r0, #1
  bne writeloop

您的原始示例正在将指针写入数组；通常称为价值等于地址".如果这是您想要的，

Your original example was writing the pointer to the array; often referred to as 'value equals address'. If this is what you want,

  ldr r0, =array_end ; finished?
  ldr r1, =array1    ; array pointer
write_loop:
  str r1, [r1], #4  ; add four and update after storing
  cmp r0, r1
  bne write_loop
; code to jump somewhere else and not execute data.

.balign 4
array1: skip 40
array_end:

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