“未定义引用‘operator>>(std::istream&, Complex&)”对于模板 Complex
我的代码：
#include <iostream>
using std::cin;
using std::cout;
using std::istream;
using std::ostream;
template<typename T>
class Complex
{
T real, img;
public:
Complex():real(0), img(0){}
friend istream& operator>>(istream& input, Complex& c1);
friend ostream& operator<<(ostream& output, Complex& c1);
Complex operator+(Complex& c1);
};
template<typename T>
istream& operator>>(istream& input, Complex<T>& c1)
{
cout<<"Real: ";
input>>c1.real;
cout<<"Imag: ";
input>>c1.img;
return input;
}
template<typename T>
ostream& operator<<(ostream& output, Complex<T>& c1)
{
output<<c1.real<<"+"<<c1.img<<"i";
return output;
}
template<typename T>
Complex<T> Complex<T>::operator+(Complex<T>& c1)
{
Complex temp;
temp.real = this->real + c1.real;
temp.img = this->img + c1.img;
return temp;
}
int main()
{
Complex<int> cmp1;
cin>>cmp1;
return 0;
}
我收到的错误位于 cin>>cmp1 ，它是 对 'operator>>(std::istream&, Complex。但我在我的代码中找不到任何问题。
如果我将使用 double 的非模板类复杂化并删除所有与模板相关的代码，则该代码有效，因此 operator>>() 的定义本质上是正确的。
当我将 Complex 设为模板时会发生什么变化？
My code:
#include <iostream>
using std::cin;
using std::cout;
using std::istream;
using std::ostream;
template<typename T>
class Complex
{
T real, img;
public:
Complex():real(0), img(0){}
friend istream& operator>>(istream& input, Complex& c1);
friend ostream& operator<<(ostream& output, Complex& c1);
Complex operator+(Complex& c1);
};
template<typename T>
istream& operator>>(istream& input, Complex<T>& c1)
{
cout<<"Real: ";
input>>c1.real;
cout<<"Imag: ";
input>>c1.img;
return input;
}
template<typename T>
ostream& operator<<(ostream& output, Complex<T>& c1)
{
output<<c1.real<<"+"<<c1.img<<"i";
return output;
}
template<typename T>
Complex<T> Complex<T>::operator+(Complex<T>& c1)
{
Complex temp;
temp.real = this->real + c1.real;
temp.img = this->img + c1.img;
return temp;
}
int main()
{
Complex<int> cmp1;
cin>>cmp1;
return 0;
}
The error I'm getting is at cin>>cmp1 which is undefined reference to 'operator>>(std::istream&, Complex<int>&)'. But I can't find anything wrong in my code.
The code works if I make complex a non-template class which uses double and remove all template-related code, so the definition of operator>>() is essentially correct.
What changes when I make Complex a template?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（2）
友元函数不是成员，因此它们不是隐式模板。那里的声明表明实例化类型
Complex存在非模板运算符。您可以使用
Friend functions are not members so they aren't implicitly templates. Declaration there suggests existence of non-template operator for instantiated type
Complex<int>.You may use
问题是目前重载的友元函数都是普通函数。如果您想将它们作为函数模板，那么您需要将类内的友元声明更改为如下所示。
更正式地说，在类模板
Complex<>的原始代码中operator<<和operator>>不是函数模板，但“普通”函数在需要时使用类模板实例化。也就是说，它们是模板化实体。有两种方法可以解决该问题，下面给出了这两种方法。
解决方案 1
演示
解决方案 2
演示
The problem is that currently the overloaded friend functions are ordinary functions. If you want to make them as function templates instead then you need to change the friend declarations inside the class to as shown below.
To be more formal, in your original code
operator<<andoperator>>for class templateComplex<>are not function templates, but “ordinary” functions instantiated with the class template if needed. That is, they are templated entities.There are 2 ways to solve the problem, both of which are given below.
Solution 1
Demo
Solution 2
Demo