将字符数组传递给外部汇编函数 [英] passing an array of chars to external assembly function

问题描述

所以我的问题很基本，但我很难在互联网上找到任何东西.假设我想用 C 编写一个函数，该函数调用用 x86_64 程序集编写的外部 nasm 函数.

so my question is basic but i had a hard time finding anything on the internet. lets say i want to write a function in C that calls an external nasm function written in x86_64 assembly.

我想向外部函数传递两个 char* 数字，对这两个数字执行一些算术运算并返回结果的 char*.我的想法是以某种方式迭代 [rdi] 和 [rsi] 并将结果保存在 rax 中(即添加 rax、[rdi]、[rsi])，但我很难真正做到这一点.遍历每个角色的正确方法是什么?增加 [rsi] 和 [rdi]?而且 - 我只需要移动到 rax 第一个字符的值对吗?

I want to pass to the external function two char* of numbers, preform some arithmetic operations on the two and return char* of the result. My idea was to iterate over [rdi] and [rsi] somehow and saving the result in rax (i.e add rax, [rdi], [rsi]) but I'm having a hard time to actually do so. what would be the right way to go over each character? increasing [rsi] and [rdi]? and also- I would only need to move to rax the value of the first character right?

提前致谢！

推荐答案

如果您可以发布汇编/C 代码 - 提出更改建议会更容易.

If you could post assembly/C code - it would be easier to suggest changes.

对于任何程序集，我都会从 C 代码开始(因为我认为在 C 中 :))然后使用编译器转换为程序集，然后根据需要在程序集中对其进行优化.假设您需要编写一个函数，该函数接受两个字符串并将它们相加并以 int 形式返回结果，如下所示:

For any assembly, I would start with a C code(since I think in C :)) and then convert to assembly using a compiler and then optimize it in the assembly as needed. Assuming you need write a function which takes two strings and adds them and returns the result as int like the following:

int ext_asm_func(unsigned char *arg1, unsigned char *arg2, int len)
{
    int i, result = 0;
    for(i=0; i<len; i++) {
        result += arg1[i] + arg2[i];
    }
    return result;
}

这是汇编(由 gcc https://godbolt.org/g/1N6vBT 生成):

Here is assembly (generated by gcc https://godbolt.org/g/1N6vBT):

ext_asm_func(unsigned char*, unsigned char*, int):
        test    edx, edx
        jle     .L4
        lea     r9d, [rdx-1]
        xor     eax, eax
        xor     edx, edx
        add     r9, 1
.L3:
        movzx   ecx, BYTE PTR [rdi+rdx]
        movzx   r8d, BYTE PTR [rsi+rdx]
        add     rdx, 1
        add     ecx, r8d
        add     eax, ecx
        cmp     r9, rdx
        jne     .L3
        rep ret
.L4:
        xor     eax, eax
        ret

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