了解开关条件下的寄存器用法 [英] Understanding register usage in the switch condition

问题描述

我有一个C语言转换条件代码和汇编代码. 但是对我来说，设置，edx或eax或ecx似乎很随意?

I have a switch-condition code in C and assembly code. But it seems very arbitrary to me what to set, edx or eax or ecx?

如何分辨edx，epx，ecx，ebp之间的区别?甚至教科书也没有给我足够的解释.

How do I tell the difference between edx, epx, ecx, ebp? Even the textbook does not give me the sufficient explanation.

#include <stdio.h>

// Enumerated type creates set of constants
// numbered 0 and upward
typedef enum {MODE_A, MODE_B, MODE_C, MODE_D, MODE_E} mode_t;

int switchmode(int *p1, int *p2, mode_t action)
{
  int result = 0;
  switch(action) {
    case MODE_A:
      result = *p1;
      *p1 = *p2;
      break;
    case MODE_B:
      *p2 += *p1;
      result = *p2;
      break;
    case MODE_C:
      *p2 = 15;
      result = *p1;
      break;
    case MODE_D:
      *p2 = *p1;
      /* Fall Through */
    case MODE_E:
      result = 17;
      break;
    default:
      result = -1;
  }
  return result;
}

int main(int argc, const char * argv[])
{
  int num1 = 10;
  int num2 = 20;

  printf("MODE_A: %d \n", switchmode(&num1, &num2, MODE_A)); // 10
  printf("MODE_B: %d \n", switchmode(&num1, &num2, MODE_B)); // 40
  printf("MODE_C: %d \n", switchmode(&num1, &num2, MODE_C)); // 20
  printf("MODE_D: %d \n", switchmode(&num1, &num2, MODE_D)); // 17
  printf("MODE_E: %d \n", switchmode(&num1, &num2, MODE_E)); // 17
  printf("NULL: %d \n", switchmode(&num1, &num2, (mode_t) NULL)); // 20
    return 0;
}

而汇编代码是

Arguments:  p1      at %ebp+8
            p2      at %ebp+12
            action  at %ebp+16

Register: result in %edx (initialized to -1)

.L17: // MODE_E
  movl $17, %edx
  jmp .L19
.L13: // MODE_A
  movl 8(%ebp), %eax
  movl (%eax), %edx
  movl 12(%ebp), %ecx
  movl (%ecx), %eax
  movl 8(%ebp), %ecx
  movl %eax, (%ecx)
  jmp .L19
.L14: // MODE_B
  movl 12(%ebp), %edx
  movl (%edx), %eax
  movl %eax, %edx
  movl 8(%ebp), %ecx
  addl (%ecx), %edx
  movl 12(%ebp), %eax
  movl %edx, (%eax)
  jmp .L19
.L15: // MODE_C
  movl 12(%ebp), %edx
  movl $15, (%edx)
  movl 8(%ebp), %ecx
  movl (%ecx), %edx
  jmp .L19
.L16: // MODE_D
  movl 8(%ebp), %edx
  movl (%edx), %eax
  movl 12(%ebp), %ecx
  movl %eax, (%ecx)
  movl $17, %edx
.L19: // default
  movl %edx, %eax // set return value

我的问题是

我怎么知道什么是eax?是论据之一吗?
如果是这样，我怎么知道哪个是eax和ecx，哪个是edx?

How do I know what is eax? Is it one of the arguments?
If so, how do I know which is the eax and ecx and which is edx?

推荐答案

在大多数汇编代码中，％eax中的内容是该函数返回的值.这就是为什么在.L19块中，％edx中的所有内容都移到％eax中的原因.往回看，无论您将result设置为什么位置，代码都会将该值移入％edx，以便在代码跳转到.L19时，默认情况下会将值放置在％eax中.
它看起来像一个32位系统，因此它具有8个通用寄存器:

In most assembly code, whatever is in %eax is the value returned from the function. That is why in the .L19 block whatever was in %edx is moved into %eax. Looking back, wherever you set result to something, the code moves that value into %edx so that when the code jumps to .L19 the value will be placed in %eax by default.
This looks like a 32-bit system, so it has 8 general-purpose registers:

• ％ebx，％ecx，％edx，％esi和％edi是寄存器，它们大部分存储在任何给定时间保存它们所需的任何数据.
• ％eax保留函数的返回值，例如调用它的函数将在更改％eax之前在％eax中查找该函数的返回值.
• ％ebp和％esp是特殊的寄存器，它们实质上管理和分配堆栈上用于各种函数调用的空间.

要查看参数的放置位置，请查看代码中数据的引用位置.例如，在.L13，程序将％ebp + 8(即p1)放入％eax，然后将该地址(* p1)处的值放入％edx.因此，％edx现在拥有p1指向的值.

寄存器不保存函数的参数.不要认为默认情况下％edx包含switchmode()的参数之一.参数总是放在堆栈上.

To see where your arguments are being placed, look at where that data is being referenced in the code. For example, at .L13, the program places %ebp+8 (which is p1) into %eax, then places the value at that address (*p1) into %edx. So, %edx now holds the value pointed to by p1.

The registers DO NOT hold the arguments to the function. Don't think that %edx contains one of the arguments to switchmode() by default. Arguments are always placed on the stack.

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