C ++指针访问类似于汇编 [英] C++ Pointer access similar to assembly

问题描述

有时我看到不同的数组访问风格在C + +，并认为它可能是相对于一个程序集寻址模式：

C ++：

  int * aa = new int [2]; 
 0 [aa] = 15; // a a different different aa [0] 
 1 [aa] = 15; 
 aa [0] = 15; 
 aa [1] = 15; 
 printf（％d％d \\\
，aa [0]，aa [1]）; 
  

装配：

  __ asm 
 {
 mov aa [0]，ebx 
 mov aa [1]，eax 
 mov 0 [aa]，ebx 
 mov 1 [aa]，eax 
} 
  

这个C ++数组访问符号a

当我尝试 [aa] 1 = 5;

/ code>，编译器给出

• 未找到aa属性，


• 缺少';'之前'常量'



• left operand必须为l-value。

  
  
  

  当我尝试指针运算时，

  
  
  

  *（aa + 1）= 0 //没有错误

  
  
  

  （aa + 0）= 0 //没有错误：）

谢谢。

解决方案

这是一个众所周知的C和C ++的怪癖，他们并不真正关心表达式的哪一半在方括号 [] 。他们都评价相同的等效表达式：

  *（0 + aa）= 15; 
  

Sometimes i see different array-access styles in C++ and thought it could be relative to an assembly addressing mode:

C++:

int * aa=new int[2];
0[aa]=15; //a little different than aa[0]
1[aa]=15;
aa[0]=15;
aa[1]=15;
printf("%d %d \n",aa[0],aa[1]);

Assembly:

__asm
{
    mov aa[0],ebx
    mov aa[1],eax
    mov 0[aa],ebx
    mov 1[aa],eax
}

Is this C++ array-access notation a standard and if yes, was it derived from an assembly addressing mode?

When i try [aa]1=5; , compiler gives

• "aa attribute not found",
• "missing ';' before 'constant'"

• "left operand must be l-value".

  //When i try pointer arithmetic,

  *(aa+1)=0//gives no error

  *(aa+0)=0//gives no error :)

Is this rule same for the operator [] overloading?

MSVC++ 2010

Thank you.

解决方案

This is a well-known quirk of C and C++, they don't really care which half of an expression goes in the square brackets []. They both evaluate to the same equivalent expression:

*(0+aa)=15;

这篇关于C ++指针访问类似于汇编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！