下面的复数解决方案是否可行？
下面看一下复数的加法操作：
#include <stdio.h>
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 Add(const Complex& p1, const Complex& p2);
};
Complex Add(const Complex& p1, const Complex& p2)
{
Complex ret;
ret.a = p1.a + p2.a;
ret.b = p1.b + p2.b;
return ret;
}
int main()
{
Complex c1(1, 2);
Complex c2(3, 4);
Complex c3 = Add(c1, c2); // c1 + c2
printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());
return 0;
}输出结果如下：
思考
Add 函数可以解决 Complex 对象相加的问题，但是 Complex 是现实世界中确实存在的复数，并且复数在数学中的地位和普通的实数相同。
为什么不能让＋操作符也支持复数相加呢？这个就涉及到操作符的重载。
C++ 中的重载能够扩展操作符的功能
操作符的重载以函数的方式进行
本质
用特殊形式的函数扩展操作符的功能
语法
下面来初探一下操作符重载：
#include <stdio.h>
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& p1, const Complex& p2);
};
Complex operator + (const Complex& p1, const Complex& p2)
{
Complex ret;
ret.a = p1.a + p2.a;
ret.b = p1.b + p2.b;
return ret;
}
int main()
{
Complex c1(1, 2);
Complex c2(3, 4);
Complex c3 = c1 + c2; // operator + (c1, c2)
printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());
return 0;
}输出结果如下：
可以将操作符重载函数定义为类的成员函数
下面来实现在成员函数中重载操作符：
#include <stdio.h>
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;
}
Complex operator + (const Complex& p)
{
Complex ret;
printf("Complex operator + (const Complex& p)\n");
ret.a = this->a + p.a;
ret.b = this->b + p.b;
return ret;
}
friend Complex operator + (const Complex& p1, const Complex& p2);
};
Complex operator + (const Complex& p1, const Complex& p2)
{
Complex ret;
printf("Complex operator + (const Complex& p1, const Complex& p2)\n");
ret.a = p1.a + p2.a;
ret.b = p1.b + p2.b;
return ret;
}
int main()
{
Complex c1(1, 2);
Complex c2(3, 4);
Complex c3 = c1 + c2; // c1.operator + (c2)
printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());
return 0;
}输出结果如下：
这个说明编译器优先在成员函数中寻找操作符重载函数
故上述代码可以直接写成：
#include <stdio.h>
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;
}
Complex operator + (const Complex& p)
{
Complex ret;
ret.a = this->a + p.a;
ret.b = this->b + p.b;
return ret;
}
};
int main()
{
Complex c1(1, 2);
Complex c2(3, 4);
Complex c3 = c1 + c2; // c1.operator + (c2)
printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());
return 0;
}
