ARM汇编循环

本文介绍了ARM汇编循环的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

限时送ChatGPT账号..

for (int i = 0; i < 10000; i++)
  a[i] = b[i] + c[i]

这种高级语言的 ARM 程序集是什么样的?

What does the ARM assembly for this high level language look like?

我还假设 A 的基址在 R8 中，基址B的地址在R9中，C的基地址在R10中和 A,B,C 都是 int 数组

I'm also assuming the base address of A is in R8, the base address of B is in R9, and the base address of C is in R10 and A,B,C are all int arrays

非常感谢

我试过了:

MOV  R0, #0  ; Init r0 (i = 0)

Loop:

        a[i] = b[i] + c[i]   //How to fix this? 

        ADD  R0, R0, #1 ;Increment it

        CMP  R0, #1000 ;Check the limit

        BLE  Loop  ;Loop if not finished

推荐答案

假设这种高级语言与 C 没有任何冲突，您可以使用 arm C 编译器从您的代码片段创建汇编代码.例如，如果您在 test.c 中有以下内容，

Assuming this high level language doesn't have anything conflicting with C, you can use an arm C compiler to create assembly code from your snippet. For example if you have the following in test.c,

void test() {
        register int i asm("r0");
        register int *a asm("r8");
        register int *b asm("r9");
        register int *c asm("r10");

        for (i = 0; i < 10000; i++) {
                a[i] = b[i] + c[i];
        }
}

你可以跑

arm-linux-androideabi-gcc -O0 -S test.c

创建一个 test.s 文件，该文件将包含您的测试函数的汇编代码以及一些额外的东西.您可以在下面看到您的循环是如何编译成程序集的.

to create a test.s file, which will contain assembly code for your test function as well as some extra stuff. You can see how your loop got compiled into to assembly below.

<snipped>
.L3:
        mov     r2, r8
        mov     r3, r0
        mov     r3, r3, asl #2
        add     r3, r2, r3
        mov     r1, r9
        mov     r2, r0
        mov     r2, r2, asl #2
        add     r2, r1, r2
        ldr     r1, [r2, #0]
        mov     ip, sl
        mov     r2, r0
        mov     r2, r2, asl #2
        add     r2, ip, r2
        ldr     r2, [r2, #0]
        add     r2, r1, r2
        str     r2, [r3, #0]
        mov     r3, r0
        add     r3, r3, #1
        mov     r0, r3
.L2:
        mov     r2, r0
        ldr     r3, .L5
        cmp     r2, r3
        ble     .L3
        sub     sp, fp, #12
        ldmfd   sp!, {r8, r9, sl, fp}
        bx      lr
<snipped>

现在这种方法的问题是相信编译器会为您的研究生成最佳代码，这可能并非总是如此，但您将得到像上述问题的快速答案，而不是等待人们:)

Now the problem with this approach is trusting the compiler generates the optimal code for your study, which might not be always the case but what you'll get is fast answers to your questions like above instead of waiting for people :)

-- 额外--

GCC 允许您将变量放入某些寄存器，请参阅 相关文档.

GCC allows you to put variables into certain registers, see related documentation.

您可以在此处获取手臂说明备忘单.

较新版本的 GCC 创建了更好的 arm 代码，正如人们所期望的那样.以上截图是由 4.4.3 版本生成的，我可以确认 Linaro 的 4.7.1 证明了我的主张.因此，如果您采用我的方法，请使用您可以获得的最新工具链.

Newer versions of GCC creates better arm code as one would expected. Above snipped is generated by version 4.4.3, and I can confirm Linaro's 4.7.1 proves my claim. So if you take my approach use the most recent tool chain you can get.

这篇关于ARM汇编循环的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！