ARM汇编中的矩阵乘法 [英] matrix multiplication in ARM assembly

问题描述

我最近在我的大学开始了一门ARM汇编课程，或者做的任务是创建一个NxM * MxP矩阵乘法程序，该程序从C代码中调用.

A course of ARM assembly recently started at my university, and or assignment is to create an NxM * MxP matrix multiplication programm, that is called from C code.

现在我对旁听者的知识非常有限，但是我非常愿意学习.我想知道的是:

Now I have a fairly limited knowledge in assambler, but i'm more than willing to learn. What I would like to know is this:

1. 如何将2D数组从C读取/传递给ASM?
2. 如何将2D数组输出回C?

我在想，我可以自己解决其余的问题，但这2点是我发现的困难.

I'm thinking, that i can figure the rest of this out by myself, but these 2 points are what I find difficult.

我在Ubuntu的qemu上的ARM汇编中使用此代码，它不在任何特定设备上使用.

I am using ARM assembly on qemu, on Ubuntu for this code, it's not going on any particular device.

推荐答案

C数组只是指针，因此当您将C数组作为assemply函数的参数传递时，您将获得一个指向内存区域的指针是数组的内容.

C arrays are merely just pointers, so when you pass a C array as an argument to an assemply function, you will get a pointer to an area of memory that is the content of the array.

要检索参数，取决于您使用的调用约定. ARM EABI 规定:

For retrieving the argument, it depends on what calling convention you use. The ARM EABI stipulates that:

前四个寄存器r0-r3(a1-a4)用于将参数值传递到子例程中并返回结果 函数的值.它们也可以用于在例程中保存中间值(但通常，仅 在子程序调用之间).

The first four registers r0-r3 (a1-a4) are used to pass argument values into a subroutine and to return a result value from a function. They may also be used to hold intermediate values within a routine (but, in general, only between subroutine calls).

对于简单函数，应根据函数签名在r0到r4中找到指向数组的指针.否则，您会在堆栈上找到它.准确找出ABI的一种好方法是反汇编调用汇编函数的C代码的目标文件，并在调用汇编函数之前检查其功能.

For simple functions, them, you should find the pointer to your array in r0 to r4 depending on your function signature. Otherwise, you will find it on the stack. A good technique to find out exactly what the ABI is would be to disassemble the object file of the C code that calls your assembly function and check what it does prior to calling your Assembly function.

例如，在Linux上，您可以在名为testasm.c的文件中编译以下C代码:

For instance, on Linux, you can compile the following C code in a file called testasm.c:

extern int myasmfunc(int *);

static int array[] = { 0, 1, 2 };

int mycfunc()
{
    return myasmfunc(array);
}

然后使用以下代码进行编译:

Then compile it with:

arm-linux-gnueabi-gcc -c testasm.c

最后使用以下命令进行拆卸:

And finally get a disassembly with:

arm-linux-gnueabi-objdump -S testasm.o

结果是:

testasm.o:     file format elf32-littlearm


Disassembly of section .text:

00000000 <mycfunc>:
   0:   e92d4800    push    {fp, lr}
   4:   e28db004    add fp, sp, #4
   8:   e59f000c    ldr r0, [pc, #12]   ; 1c <mycfunc+0x1c>
   c:   ebfffffe    bl  0 <myasmfunc>
  10:   e1a03000    mov r3, r0
  14:   e1a00003    mov r0, r3
  18:   e8bd8800    pop {fp, pc}
  1c:   00000000    andeq   r0, r0, r0

您可以看到通过将参数放入寄存器r0中来调用单参数函数myasmfunc. ldr r0, [pc, #12]的含义是将pc+12处的内存地址的内容加载到r0中".那就是指向数组的指针的存储位置.

You can see that the single-parametered function myasmfunc is called by putting the parameter into register r0. The meaning of ldr r0, [pc, #12] is "load into r0 the content of the memory address that is at pc+12". That is where the pointer to the array is stored.

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