运算符重载的概念是什么 [英] what is the concept of operator overloading

问题描述

我已经用TC编译器编写了一个程序..并出现错误，但无法理解

I have written a program in TC compiler..and getting an error but could not understand

#include<iostream.h>
#include<conio.h>
class aman
{
	int real;
	int imag;
	public : aman(int a,int b)
		{
			real=a;
			imag=b;
		}
	       void showdata()
		{
			cout<<real<<"+i"<<imag<<endl;
		}
		 aman operator+(aman);
};
       aman aman::operator + (aman c)
	      {

		aman temp;

		temp.real=real+ c.real;
		temp.imag=imag+ c.imag;
		return (temp);
	      }
void main()
{
	clrscr();
	aman d,e,f;
       d=aman (5,6);
       e= aman (6,9);
	f=d+e;
	d.showdata();
	cout<<"a is";
	e.showdata();
	cout<<"b is";
	f.showdata();

	cout<<"f=d+e is:"<<endl;
	f.showdata();
	getch();
}

Error: could not find match for ‘aman::aman()’

它指向

and it is pointing on

aman aman::operator + (aman c)
	      {

		aman temp;
		temp.real=real+ c.real;
		temp.imag=imag+ c.imag;
		return (temp);
	      }

在此先感谢

Thanks in advance

推荐答案

一旦提供了参数化的构造函数，编译器将不会为您提供默认的普通构造函数.您需要手动定义普通ctor.

aman :: aman()
{
}

Once you provide a parameterized constructor, the compiler doesnt provide you the default plain one. You need to define your plain ctor manually.

aman::aman()
{
}

您还应该通过引用或指针将其传递给+运算符函数.
默认值是按值传递，这就是这里发生的情况.这将创建变量的完整副本，然后使用该副本调用函数，然后删除该副本.通过引用传递和通过指针传递只是传递内存地址的副本.这更快，特别是对于包含大量数据的类

You should also be passing by reference or pointer into the + operator function.
Pass by value is the default, and is what is happening here. This creates a full copy of the variable, then calls the function with this copy, then deletes the copy. Pass by reference and pass by pointer just pass a copy of the memory address. This is quicker, especially for classes which hold lots of data

class aman {
	//other functions here
	aman operator +(const aman&);
};

aman aman::operator +(const aman &c) {
	aman temp;
	temp.real = real + c.real;
	temp.imag = imag + c.imag;
	return (temp);
}

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