C++运算符重载(5) 重载== explicit避免隐式转换

有的时候程序中存在隐藏式转换

比如下面这个例子

#include<iostream>using namespace std;class myComplex {private:int real;   //复数的实部int image;  //复数的虚部public:myComplex(int real = 0, int image = 0) {this->real = real;this->image = image;}friend bool operator == (const myComplex &obj1, const myComplex &obj2);};bool operator == (const myComplex &obj1, const myComplex &obj2) {if (obj1.real == obj2.real && obj1.image == obj2.image)return true;return false;}int main() {myComplex TestA(3,0);myComplex TestB(10,0);if (TestA == 3) {cout << "SAME" << endl;}else {cout << "NOT SAME" << endl;}system("PAUSE");return 0;

我们通过重载操作符 == 来判断两个自定义的类:复数是否相等

myComplex(int real = 0, int image = 0)

在构造函数中,我们给形参定义了默认值 所以在生成对象时 如果不指定形参 便会使用默认值

C++中,如果构造函数可以只传入一个参数来调用,则会发生隐式转换

if (TestA == 3)

在上面行代码中 将3传入重载操作符函数 构造函数便会构造一个3,0的myComplex对象,这便发生了隐式转换(将int型转为myComplex型)

我们可以通过使用explicit来避免隐式转换

explicit myComplex(int real = 0, int image = 0) {this->real = real;this->image = image;}

这时必须指明类型

myComplex TestA(3,0);myComplex TestB(10,0);if (TestA == (myComplex)3) {    //必须指明类型cout << "SAME" << endl;}else {cout << "NOT SAME" << endl;}