问题描述
阅读后这篇文章在 Herb Sutter 的博客上，我进行了一些实验，但遇到了一些令我感到困惑的事情.我使用的是 Visual C++ 2005，但如果这取决于实现，我会感到惊讶.
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.
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，它显示~Base()"两次...但是如果我取消注释构造函数，它只显示一次！
When run, it displays "~Base()" twice... But if I un-comment the constructor, it displays it only once!
有人对此有解释吗?
推荐答案
这取决于实现.
该标准允许在将临时引用绑定到常量引用时进行复制.在您的情况下，VC++ 仅在隐式定义构造函数时执行复制.这是出乎意料的，但也是允许的.
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 将解决这个问题.
这篇关于对临时的常量引用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跟版网！