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C++语言基础(22)-转换构造函数和类型转换函数

时间:2017-07-05 21:16:11      阅读:168      评论:0      收藏:0      [点我收藏+]

标签:转换函数   eal   int   导致   style   div   public   使用   double   

一.转换构造函数

将其它类型转换为当前类类型需要借助转换构造函数(Conversion constructor)。转换构造函数也是一种构造函数,它遵循构造函数的一般规则。转换构造函数只有一个参数。

#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){ }
    Complex(double real): m_real(real), m_imag(0.0){ }  //转换构造函数
public:
    friend ostream & operator<<(ostream &out, Complex &c);  //友元函数
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;
}

int main(){
    Complex a(10.0, 20.0);
    cout<<a<<endl;
    a = 25.5;  //调用转换构造函数
    cout<<a<<endl;
    return 0;
}

运行结果:

10 + 20i
25.5 + 0i

二.类型转换函数

#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:

f = 24.6
f = 43.1
n = 43

注意:最好不要同时使用转换构造函数和类型转换函数,因为这样会导致语义的二义性。

 

C++语言基础(22)-转换构造函数和类型转换函数

标签:转换函数   eal   int   导致   style   div   public   使用   double   

原文地址:http://www.cnblogs.com/yongdaimi/p/7123124.html

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