operator type(){
//TODO:
return data;
}
#include <iostream>
using namespace std;
//复数类
class Complex{
public:
Complex(): m_real(0.0), m_imag(0.0){ }
Complex(double real, double imag): m_real(real), m_imag(imag){ }
public:
friend ostream & operator<<(ostream &out, Complex &c);
friend Complex operator+(const Complex &c1, const Complex &c2);
operator double() const { return m_real; } //类型转换函数
private:
double m_real; //实部
double m_imag; //虚部
};
//重载>>运算符
ostream & operator<<(ostream &out, Complex &c){
out << c.m_real <<" + "<< c.m_imag <<"i";;
return out;
}
//重载+运算符
Complex operator+(const Complex &c1, const Complex &c2){
Complex c;
c.m_real = c1.m_real + c2.m_real;
c.m_imag = c1.m_imag + c2.m_imag;
return c;
}
int main(){
Complex c1(24.6, 100);
double f = c1; //相当于 double f = Complex::operator double(&c1);
cout<<"f = "<<f<<endl;
f = 12.5 + c1 + 6; //相当于 f = 12.5 + Complex::operator double(&c1) + 6;
cout<<"f = "<<f<<endl;
int n = Complex(43.2, 9.3); //先转换为 double,再转换为 int
cout<<"n = "<<n<<endl;
return 0;
}
运行结果:
operator double() const { return m_real; } //转换为double类型
operator int() const { return (int)m_real; } //转换为int类型
那么下面的写法就会引发二义性:
Complex c1(24.6, 100); float f = 12.5 + c1;编译器可以调用 operator double() 将 c1 转换为 double 类型,也可以调用 operator int() 将 c1 转换为 int 类型,这两种类型都可以跟 12.5 进行加法运算,并且从 Complex 转换为 double 与从 Complex 转化为 int 是平级的,没有谁的优先级更高,所以这个时候编译器就不知道该调用哪个函数了,干脆抛出一个二义性错误,让用户解决。
版权说明:Copyright © 广州松河信息科技有限公司 2005-2025 版权所有 粤ICP备16019765号
广州松河信息科技有限公司 版权所有