这个C ++语法是什么意思，为什么它工作？ [英] What does this C++ syntax mean and why does it work?

问题描述

我正在浏览OpenDE的源代码，我发现了一个类的数组索引操作符'[]'的一些奇怪的语法用法。下面是一个简化的示例来显示语法：

  #include< iostream& 
 
 class Point 
 {
 public：
 Point（）：x（2.8），y（4.2），z（9.5）{} 
 
 operator const float *（）const 
 {
 return& x; 
} 
 
 private：
 float x，y，z; 
}; 
 
 int main（）
 {
 Point p; 
 std :: cout<< x：<< p [0]<< '\\\
'
<< y：< p [1]<< '\\\
'
<< z：<< p [2]; 
} 
  

输出：

  x：2.8 
y：4.2 
z：9.5 
  

$ b b

这里发生了什么？为什么这种语法工作？ Point类不包含重载的 operator [] ，这里的代码试图自动转换为float。

感谢

div

被隐式转换为 const float * const ，它指向 x 。所以 * p 是 x ， *（p + 1） y ，等等。
当然，这是一个很奇怪的想法（和令人困惑的！）这样做。通常最好在数组中存储x，y和z，并且有一个函数来获取整个数组，如果他们真的想这样做。

I was looking through the source of OpenDE and I came across some wierd syntax usage of the array indexing operator '[]' on a class. Here's a simplified example to show the syntax:

#include <iostream>

class Point
{
public:
    Point() : x(2.8), y(4.2), z(9.5) {}

    operator const float *() const
    {
        return &x;
    }

private:
    float x, y, z;
};

int main()
{
    Point p;
    std::cout << "x: " << p[0] << '\n'
              << "y: " << p[1] << '\n'
              << "z: " << p[2];
}

Output:

x: 2.8
y: 4.2
z: 9.5

What's going on here? Why does this syntax work? The Point class contains no overloaded operator [] and here this code is trying to do an automatic conversion to float somewhere.

I've never seen this kind of usage before -- it definitely looks unusual and surprising to say the least.

Thanks

解决方案

p is being converted implicitly into a const float* const, which points to x. So *p is x, *(p+1) is y, and so on. It's a pretty weird idea (and confusing!) to do it this way, of course. It's usually preferable to store x, y, and z in an array and have a function to get the entire array if they really want to do things this way.

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