一、操作符重载:
1、我们先来看一个问题实现,下面的复数解决方案是否可行,复数大家应该都不陌生(分为实部和虚部):
class Complex
{
public:
int a;
int b;
};
/*--------------------------*/
int main()
{
Complex c1 = {1,2};
Complex c2 = {3,4};
Complex c3 = c1 + c2;//也就是复数的实部加实部,虚部叫虚部。
return 0;
}
注:成员变量为公有且没有自定义构造函数的时候,可以通过大括号来分别初始 化成员变量;
代码版本一:
#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;
}
运行结果:
root@txp-virtual-machine:/home/txp# ./a.out
c3.a=4,c3,b=6
这里通过Add函数可以解决Complex对象相加的问题,但是在我们数学运算里面就是直接实部加实部,虚部加虚部,和正常的实数相加一样,所以说,为什么不直接这样操作呢,这就涉及到符号"+"的问题。
2、操作重载符的引出
3、操作重载符的语法:
Type operator Sign(const Type& p1,const Type& p2)
{
Type ret;
return ret;
}
注:Sign为系统中预定义的操作符,比如:+、-、*、/ 等
代码版本二:
#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:
root@txp-virtual-machine:/home/txp# ./a.out
c3.a=4,c3,b=6
4、再次改进代码:
class Type
{
public:
Type operator Sign(const Type& p)
{
Type ret;
return ret;
}
};
代码版本三:
#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:
root@txp-virtual-machine:/home/txp# ./a.out
Complex operator +(const Complex& p)
c3.a=4,c3,b=6
二、总结:
好了,今天的分享就到这里,如果文章中有错误或者不理解的地方,可以交流互动,一起进步。我是txp,下期见!