Const引用临时 [英] Const reference to temporary

问题描述

阅读本文在Herb Sutter的博客上，我尝试了一下，遇到了困惑我的东西。我使用Visual C ++ 2005，但如果这是实现依赖的，我会很惊讶。

这是我的代码：

  #include< iostream> 
 
 using namespace std; 
 
 struct Base {
 // Base（）{} 
〜Base（）{cout< 〜Base（）<< endl; } 
}; 
 
 int main（）
 {
 const Base& f = Base（）; 
} 
  

运行时会显示〜Base 两次 ...但如果我取消注释构造函数，它只会显示一次！

$

这个IS实现依赖。 >

标准允许在将临时数据绑定到const引用时发生复制。在你的情况下，VC ++只在构造函数被隐式定义时才执行一个复制。这是意外的，但允许。

C ++ 1x 将修复此问题。

After reading this article on Herb Sutter's blog, I experimented a bit and ran into something that puzzles me. I am using Visual C++ 2005, but I would be surprised if this was implementation dependent.

Here is my code:

#include <iostream>

using namespace std;

struct Base {
    //Base() {}
    ~Base() { cout << "~Base()" << endl; }
};

int main()
{
    const Base & f = Base();
}

When run, it displays "~Base()" twice... But if I un-comment the constructor, it displays it only once!

Does anyone have an explanation for this?

解决方案

This IS implementation dependent.

The standard allows a copy to occur when binding a temporary to a const reference. In your case, VC++ performs a copy only when the constructor is implicitly defined. This is unexpected, but permitted.

C++1x will fix this.

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