C++操作符重载！

操作符重载为操作符提供不同的语义

#include <iostream>

using namespace std;

struct Complex
{
    int a;
    int b;
};
int main()
{
    Complex c1 = {1,2};
    Complex c2 = {3,4};
    Complex c3 = c1 + c2;//编译出错

    cout << "Press any key to continue..." << endl;
    cin.get();
    return 0;
}
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上个例子中，不能用“+”直接操作两个结构体，改动一下，提供一个函数。

//增加一个add函数
#include <iostream>

using namespace std;

struct Complex
{
    int a;
    int b;
};
Complex add(const Complex& c1,const Complex& c2)
{
    Complex ret;
    ret.a = c1.a + c2.a;
    ret.b = c1.b + c2.b;
    return ret;
}
int main()
{
    Complex c1 = {1,2};
    Complex c2 = {3,4};
    Complex c3 = add(c1,c2);//编译出错
    cout<<"c3.a = "<<c3.a<<endl;
    cout<<"c3.b = "<<c3.b<<endl;

    cout << "Press any key to continue..." << endl;
    cin.get();
    return 0;
}
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C++中操作符重载的本质

C++中通过operator关键字可以利用函数拓展操作符，operator的本质是通过函数重载实现操作符重载。

//把add替换为“operator+"
#include <iostream>

using namespace std;

struct Complex
{
    int a;
    int b;
};
Complex operator+ (const Complex& c1,const Complex& c2)
{
    Complex ret;
    ret.a = c1.a + c2.a;
    ret.b = c1.b + c2.b;
    return ret;
}

int main()
{
    Complex c1 = {1,2};
    Complex c2 = {3,4};
    Complex c3 = operator+ (c1,c2);
    cout<<"c3.a = "<<c3.a<<endl;
    cout<<"c3.b = "<<c3.b<<endl;
    c3 = c1 + c2;
    cout<<"c3.a = "<<c3.a<<endl;
    cout<<"c3.b = "<<c3.b<<endl;
    cout << "Press any key to continue..." << endl;
    cin.get();
    return 0;
}
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operator关键字拓展的操作符应用到类。

利用友元friend关键字可以例外的开放private声明的类成员的使用权限

//友元friend和全局函数 使操作符重载应用与类
#include <iostream>

using namespace std;

class Complex
{
    int a;
    int b;
public:
    Complex(int a=0,int b=0)
    {
        this->a = a;
        this->b = b;
    }
    int getA()
    {
        return a;
    }
    int getB()
    {
        return b;
    }
    friend Complex operator+ (const Complex& c1,const Complex& c2);
};
Complex operator+ (const Complex& c1,const Complex& c2)
{
    Complex ret;
    ret.a = c1.a + c2.a;
    ret.b = c1.b + c2.b;
    return ret;
}

int main()
{
    Complex c1(1,2);
    Complex c2(3,4);
    Complex c3 = c1 + c2;
    cout<<"c3.a = "<<c3.getA()<<endl;
    cout<<"c3.b = "<<c3.getB()<<endl;
    cout << "Press any key to continue..." << endl;
    cin.get();
    return 0;
}
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以上是“+”操作符，接下来重载“<<”操作符

//左移操作符重载
#include <iostream>

using namespace std;

class Complex
{
    int a;
    int b;
public:
    Complex(int a=0,int b=0)
    {
        this->a = a;
        this->b = b;
    }
    int getA()
    {
        return a;
    }
    int getB()
    {
        return b;
    }
    friend Complex operator+ (const Complex& c1,const Complex& c2);
    friend ostream& operator<< (ostream& out,const Complex& c);
};
Complex operator+ (const Complex& c1,const Complex& c2)
{
    Complex ret;
    ret.a = c1.a + c2.a;
    ret.b = c1.b + c2.b;
    return ret;
}
ostream& operator<< (ostream& out,const Complex& c)
{
    out<<c.a<<"+"<<c.b<<"i";
    return out;//返回ostream类型的out 是为了能连续输出
}
int main()
{
    Complex c1(1,2);
    Complex c2(3,4);
    Complex c3 = c1 + c2;
    //cout<<c1; -> operator<<(cout,c1) 如果没有返回cout的话，不能接着输出endl
    cout<<c1<<endl;//->((operator<<(cout,c1)))<<endl;
    cout<<c2<<endl;
    cout<<c3<<endl;
    cout << "Press any key to continue..." << endl;
    cin.get();
    return 0;
}
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小结:

1、操作符重载的本质是通过函数扩展操作符的语义    2、operator关键字是操作符重载的关键

3、friend关键字可以函数或类开发访问权限               4、操作符重载遵循函数重载的规则

通过operator关键字能够讲操作符定义为全局函数，操作符重载的本质就是函数重载，那么类的成员函数是否可以作为操作符重载的函数？

#include <iostream>
#include <cstdlib>
using namespace std;

using namespace std;

class Complex
{
    int a;
    int b;
public:
    Complex(int a,int b)
    {
        this ->a = a;
        this ->b = b;
    }
    int getA()
    {
        return a;
    }
    int getB()
    {
        return b;
    }
    Complex operator+ (const Complex& c2);
    friend ostream& operator<< (ostream& out,const Complex& c);
};
Complex Complex::operator+ (const Complex& c2)
{
    Complex ret(0,0);
    ret.a = this->a + c2.a;
    ret.b = this->b + c2.b;
    return ret;
}
ostream& operator<< (ostream& out,const Complex& c)
{
    out<<c.a<<" + "<<c.b<<"i";//数学里的虚数表示法
    return out;//考虑c++标准库重载的左移操作符支持链式调用，所以要返回cout,如果不返回，则不能接着返回endl了；
}
int main(int argc,char *argv[])
{
    Complex c1(1,2);
    Complex c2(3,4);
    Complex c3 = c1 + c2;//->c3 = c1.operator+(c2)

    cout<<c1<<endl;
    cout<<c2<<endl;
    cout<<c3<<endl;
    cout << "Press the enter key to continue ...";
    cin.get();
    return EXIT_SUCCESS;
}
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输出结果一样。

用成员函数重载的操作符比全局操作符函数少一个参数，即左操作符，而且不需要friend。

问题来了，什么时候用全局的函数重载，什么时候有成员函数重载。

1，当无法修改左操作数的类时，使用全局函数进行重载  2，=,[],(),->操作符只能通过成员函数进行重载。

//Array.h
#ifndef ARRAY
#define ARRAY
class Array
{
private:
    int mLength;
    int* mSpace;
public:
    Array(int length);
    Array(const Array& obj);
    int length();
    ~Array();
    int& operator[] (int i);
    Array& operator= (const Array& obj);
    bool operator== (const Array& obj);
    bool operator!= (const Array& obj);
};
#endif // ARRAY
//Array.c
#include "array.h"

Array::Array(int length)
{
    if(length < 0)
    {
        length = 0;
    }
    mLength = length;
    mSpace = new int[mLength];
}
Array::Array(const Array &obj)
{
    mLength = obj.mLength;
    mSpace = new int[mLength];
    for(int i = 0;i<mLength;i++)
    {
        mSpace[i] = obj.mSpace[i];
    }
}
int Array::length()
{
    return mLength;
}
Array::~Array()
{
    mLength = -1;
    delete[] mSpace;
}
int& Array::operator[] (int i)//作为左值的调用语句返回的必须是引用
{
    return mSpace[i];
}
Array& Array::operator= (const Array& obj)//返回Array的引用是为了可以连续赋值
{                                //a3 = a2 = a1;->a3 = a2.operator=(a1);
    delete[] mSpace;//先释放自己的原有的堆空间 因为接下来要申请
    mLength = obj.mLength;
    mSpace = new int[mLength];
    for(int i = 0;i<mLength;i++)
    {
        mSpace[i] = obj.mSpace[i];
    }
    return *this;
}
bool Array::operator== (const Array& obj)
{
    bool ret = true;
    if(mLength==obj.mLength)
    {
        for(int i=0;i<mLength;i++)
        {
            if(mSpace[i] != obj.mSpace[i])
            {
                return false;
                break;
            }
        }
    }
    else
    {
        ret = false;
    }
    return ret;
}
bool Array::operator!= (const Array& obj)
{
    return !(*this == obj);
}
//main.c
#include <iostream>
#include <cstdlib>
#include "array.h"
using namespace std;

int main(int argc,char *argv[])
{
    Array a1(10);
    Array a2(0);
    Array a3(1);
    for(int i = 0;i<a1.length();i++)
    {
        a1[i] = i+  1;
    }

   for(int i = 0;i < a1.length();i++)
    {
        cout<<"Element"<<i<<":"<<a1[i]<<endl;
    }

     a2 = a1;//-> a2.operator=(a1);
     //a3 = a2 = a1;//->a3 = a2.operator=(a1);
    for(int i=0;i<a2.length();i++)
    {
         cout<<"Element"<<i<<":"<<a2[i]<<endl;
    }
    if(a1 == a2)
    {
        cout<<"a1 == a2"<<endl;
    }
    if(a3!=a2)
    {
        cout<<"a3 != a2"<<endl;
    }
    cout << "Press the enter key to continue ...";
    cin.get();
    return 0;
}
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为什么要用到赋值操作符重载？

C++编译器会为每个类提供默认的赋值操作符，但默认的赋值操作符只是简单的值复制，一旦有指针类的成员变量则就只复制指针，它们将指向同一片空间，而相应的空间就没用了，所以类中存在指针成员变量时就需要重载赋值操作符

++操作符的重载

++操作符只有一个操作数，且有前缀和后缀的区分。如何重载++操作符才能区分前置运算和后置运算呢？

#include <iostream>
#include <cstdlib>
using namespace std;

using namespace std;

class Complex
{
    int a;
    int b;
public:
    Complex(int a,int b)
    {
        this ->a = a;
        this ->b = b;
    }
    int getA()
    {
        return a;
    }
    int getB()
    {
        return b;
    }
    Complex operator+ (const Complex& c2);
    Complex operator ++(int);//占位参数
    Complex& operator ++();
    friend ostream& operator<< (ostream& out,const Complex& c);
};
Complex Complex::operator+ (const Complex& c2)
{
    Complex ret(0,0);
    ret.a = this->a + c2.a;
    ret.b = this->b + c2.b;
    return ret;
}
ostream& operator<< (ostream& out,const Complex& c)
{
    out<<c.a<<" + "<<c.b<<"i";//数学里的虚数表示法
    return out;//考虑c++标准库重载的左移操作符支持链式调用，所以要返回cout,如果不返回，则不能接着返回endl了；
}
Complex Complex:: operator ++(int)
{
    Complex ret = *this;//先将当前对象做一个备份
    /*首先解释++后缀的原理
     int a=0;
     cout<<a++<<endl;则会打印0 且a=1。
     为了模拟这个过程则要先保存一个当前值的备份
    */
    a++;//然后当前对象进行+1操作
    b++;
    return ret;//将备份返回
}
Complex& Complex:: operator ++()//为什么要返回引用？
{
    ++a;
    ++b;
    return *this;
}
int main(int argc,char *argv[])
{
    Complex c1(1,2);
    Complex c2(3,4);
    Complex c3 = c1 + c2;//->c3 = c1.operator+(c2)

    cout<<c1<<endl;
    cout<<c2<<endl;
    cout<<c3<<endl;
    cout << "Press the enter key to continue ...";
    cin.get();
    return EXIT_SUCCESS;
}
View Code

为什么不用重载&&和||操作符？

首先在语法上是没有错误的，但最好不要去重载，&&和||的特性是短路。

//短路
int a1 = 0;
int a2 = 1;
if(a1&&(a1+a2))
{
    cout<<"Hello"<<endl;
}
//因为a1=0，则(a1+a2)根本不会执行。这个就是所谓的短路。
//如果重载了&&
class Test
{
    int a;
public:
    Test(int i)
    {
      this->a = i;
    }
    Test operator+ (const Test& obj) 
   {
      Test ret = 0;
       ret.a = this->a+obj.a;
       return ret;
    }
    bool operator&& (const Test& obj)
    {
        return a&&obj.a;
    }
};
Test t1 = 0;
Test t2 =1;
if(t1.operator&&(t1.operator+(t2)))
{
    cout<<"Hello"<<endl;
}
//则会先执行+操作,违反了短路原则。

&&和||是c++中非常特殊的操作符。

&&和||内置实现了短路规则，操作符重载是靠函数重载来实现的，操作数作为函数参数传递，c++的函数参数都会被求值，无法实现短路规则。

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原文地址：https://www.cnblogs.com/Rainingday/p/7574223.html