ARM汇编-访问参数vs返回值？ [英] ARM assembly - access parameter vs return value?

问题描述

我有一个函数原型 int Palindrome（const char * c_style_string）;

在ARM v8程序集中，我相信该参数存储在寄存器w0中。但是，这不是 ret 输出值的寄存器吗？

In ARM v8 assembly, I believe that the parameter is stored in register w0. However, isn't this also the register that ret outputs the value of?

如果是，我该怎么办？需要这样做以确保值不会被覆盖吗？我在代码的开头就想到了 mov w0，w1 这样的东西，以便每次解析它时都将c_style_string称为w1，然后编辑w0以存储一个int ...这对吗？

If so, what do I need to do so that values do not get overwritten? I was thinking something like mov w0, w1 at the beginning of my code so that I refer to c_style_string as w1 whenever I parse through it, and then edit w0 to store an int...would this be right?

谢谢！

推荐答案

您可能要按照用于ARM 64位体系结构的ABI 。

在上面的示例中，您可以将c_style_string的地址保留在被调用者保存的寄存器（X19-X29）中，并在每次调用回文式时将其复制到x0 / w0（ ）-我在这里假设Palindrome（）是C函数，因此它本身符合ARCH 64位ABI。

一个理想的副作用是您的C代码可以始终调用您的汇编代码，反之亦然。

You may want to write your assembly code in compliance with the ABI for ARM 64-bit Architecture.
In the example above, you could keep the address for c_style_string in a 'Callee-saved' register (X19-X29)', and copy it to x0/w0 every time you are calling a Palindrome() - I am assuming here Palindrome() is a C function, and is therefore itself compliant with the ARCH 64-bit ABI.
A desirable side-effect would be that your C code could call always your assembly code, and vice-versa.

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