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运算符重载（二）
● 对称运算符通常定义为非成员函数以支持首个操作数的类型转换

struct Str
{int val = 0;Str(int input): val(input){}auto operator+(Str x){std::cout << "auto operator+(Str x)\n";return Str(val + x.val);}
};
int main()
{Str x = 3;Str z = x + 4; //通过类Str的构造函数将4转换为Str类型与x相加Str u = 4 + x; //Error: Invalid operands to binary expression ('int' and 'Str')return 0;
}

struct Str
{int val = 0;Str(int input): val(input){}
};
auto operator+(Str input1, Str input2)
{std::cout << "auto operator+(Str input1, Str input2)\n";return Str(input1.val + input2.val);
}
int main()
{Str x = 3;Str y = x + 3; OKreturn 0;
}

struct Str
{Str(int input): val(input){}friend auto operator+(Str, Str);
private:int val = 0;
};
auto operator+(Str input1, Str input2)
{std::cout << "auto operator+(Str input1, Str input2)\n";return Str(input1.val + input2.val);
}
int main()
{Str x = 3;Str y = x + 3;Str z = 3 + y;return 0;
}

● 移位运算符一定要定义为非成员函数，因为其首个操作数类型为流类型

struct Str
{Str(int input): val(input){}friend auto operator+(Str input1, Str input2){return Str(input1.val + input2.val);}friend auto& operator<<(std::ostream& output, Str input) //返回引用，第一个参数类型是std::ostream{output << input.val;return output;}
private:int val = 0;
};
int main()
{Str x = 3;Str y = x + 3;std::cout << x << ' ' << y <<std::endl;return 0;
}

● 赋值运算符也可以接收一般参数

struct Str
{Str(): val(0){}Str(int input): val(input){}Str& operator=(const Str& input) //重载赋值运算符1。重载运算符只接受一个参数，因为缺省参数是*this。{val = input.val;return *this;}Str& operator=(const std::string& input) //重载赋值运算符2{val = static_cast<int>(input.size());return *this;}
//private:int val;
};
int main()
{Str x = 3;//Str y = "12345"; //调用构造函数而非重载赋值运算符Str y;y = "12345"; //调用重载赋值运算符std::cout << y.val << std::endl;return 0;
}

● operator [] 通常返回引用

struct Str
{Str(): val(0){}Str(int input): val(input){}Str& operator=(const Str& input) //重载赋值运算符1，{val = input.val;return *this;}Str& operator=(const std::string& input) //重载赋值运算符2{val = static_cast<int>(input.size());return *this;}//int operator[](int id) //返回“值”，只读，不可执行写操作int& operator[](int id) //#1返回引用，可读可写，但是*this可被修改{return val;}int operator[](int id) const //#2const修饰构成重载返回“值”，可读不可写，即*this不可被修改{return val;}
//private:int val;
};
int main()
{Str x = 3;x = "12345";std::cout << x[0] << '\n'; //读x[0] = 100; //写std::cout << x[0] << '\n';const Str cx = 3;//std::cout << cx[0] << std::endl; //见#1,Error: No viable overloaded operator[] for type 'const Str'std::cout << cx[0] << std::endl; //OK，见#2return 0;
}

● 自增、自减运算符的前缀、后缀重载方法

struct Str
{Str(): val(0){}Str(int input): val(input){}//Str operator++() //返回“值”，Error: cannot increment value of type 'void'Str& operator++() //前缀自增{++val;return *this;}Str operator++(int) //后缀自增,返回“值”{Str tmp(*this); //调用拷贝构造，构造临时对象，编译器不一定能优化，导致性能上的损失++val;return tmp;}
//private:int val;
};
int main()
{Str s;++(++s); //调用前缀自增std::cout << s.val <<std::endl;std::cout << (s++).val << std::endl; //调用后缀自增std::cout << s.val <<std::endl;return 0;
}

● 使用解引用运算符（ * ）与成员访问运算符（ -> ）模拟指针行为
– 注意“ .” 运算符不能重载

通过类对象而不是指向类对象的指针调用其成员的，所以不能重载

struct Str
{Str(int* p): ptr(p){}//operator*() //Error: C++ requires a type specifier for all declarationsint& operator*() //返回引用，支持读写操作{return *ptr;}int operator*() const //返回“值”，只读{return *ptr;}
private:int* ptr;
};
int main()
{int x = 100;Str ptr(&x);std::cout << *ptr << std::endl; //读*ptr = 101; //写std::cout << *ptr << std::endl;return 0;
}

– “→” 会递归调用 操作 “→”

struct Str
{Str(int* p): ptr(p){}Str* operator ->() //重载运算符本质上是个函数{return this;}int val = 5;
private:int* ptr;
};
int main()
{int x = 100;Str ptr(&x);std::cout << ptr->val << std::endl; //#1std::cout << (ptr.operator->()->val) << std::endl; //#2 #1与#2等价return 0;
}

struct Str2
{Str2* operator->(){return this;}int blabla = 20;
};struct Str
{Str(int* p): ptr(p){}Str2 operator ->() //类Str中重载->运算符返回Str2类对象{return Str2{};}int val = 5;
private:int* ptr;
};
int main()
{int x = 100;Str ptr(&x);std::cout << ptr->blabla << std::endl; //#1std::cout << (ptr.operator->().operator->()->blabla) << std::endl; //#2 #1与#2等价return 0;
}

int operator->() //Error: member type 'int' is not a pointer
{return blabla;
}int* operator->() //Error: member reference base type 'int' is not a structure or a union
{return &blabla;
}

● 使用函数调用运算符构造可调用对象

struct Str
{Str(int p): val(p){}int operator()(){return val;}int operator()(int x, int y, int z) //参数列表不同，重载{return val + x + y +z;}
private:int val;
};
int main()
{ Str obj(100);std::cout << obj() << std::endl;std::cout << obj(1, 2, 3) << std::endl;return 0;
}

struct Str
{Str(int p): val(p){}int& operator()(){return this->val;}bool operator()(int input) //参数依据实际情况修改，更加灵活，是Lambda表达式的基础{//return val < input;return val++ < input;}
private:int val;
};
int main()
{ Str obj(100);std::cout << obj() << std::endl;std::cout << obj(1) << std::endl;std::cout << obj() << std::endl;std::cout << obj(199) << std::endl;std::cout << obj() << std::endl;return 0;
}

参考
深蓝学院：C++基础与深度解析