T t2(t1) 等价于 T t2 = t1 么-优快云博客

本文探讨了C++中显式构造函数的作用及其如何避免隐式类型转换，通过实例对比了使用与未使用explicit关键字的区别。

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Let's see the code:

// test.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
using namespace std;

class T
{
public:
    T(){}
    ~T(){}
    T(const T &other)
    {
        cout <<"copy constructor" <<endl;
    }
    T &operator=(const T &other)
    {
        cout <<"operator=()" <<endl;
return *this;    
    }
};

int _tmain(int argc, _TCHAR* argv[])
{
    T t;

    T t1(t);
    T t2 = t;

return 0;
}

As we all know, 't1' and 't2' will call copy constructor to initialize the new object. But if I change the code as follow:

class T
{
public:
    T(){}
    ~T(){}
explicit T(const T &other)        // Here we add keyword 'explicit' to avoid implicit convertion.
    {
        cout <<"copy constructor" <<endl;
    }
    T &operator=(const T &other)
    {
        cout <<"operator=()" <<endl;
return *this;    
    }
};

int _tmain(int argc, _TCHAR* argv[])
{
    T t;

    T t1(t);            // Explicit convertion. Ok.
    T t2 = t;           // Implicit convertion. Compilation error!

return 0;
}