朋友函数在模板定义

问题描述

我的问题ist与有关。

我想重载运算符<<对于一些类，我发现两个不同的符号都工作：

  c>  code>用于特定类型 A< T> ，而第二种使任何 
 
 
 
因此，是的，第一个更具限制性：
  template< class T> 
 ostream&运算符<< （ostream& os，const A< T& a）{
 A double> b（0.0）; 
 b.t; //编译错误版本1，精细版本2 
 return os; 
} 
 
 int main（）{
 A< int> a（0）; 
 cout<< a<< endl; 
} 
  
 
My question ist related a bit to this one.
I want to overload the operator << for some class and I found two different notations that both work:
template <class T>
class A{
  T t;
public:
  A(T init) : t(init){}
  friend ostream& operator<< <> (ostream &os, const A<T> &a); //need forward declaration
  //template <class U> friend ostream& operator<< (ostream &os, const A<U> &a);
};
Do I define identical things with different notations? Or is the first version more restrictive in which instance (in this case only the instance with the same T as my class A) of << is friend of A?
 解决方案 
The first version restricts the friendship to the operator<< for the specific type A<T> , while the second makes any operator<< that takes an A<SomeType> a friend.
So yes, the first one is more restrictive:
template<class T>
ostream& operator<< (ostream& os, const A<T>& a) {
    A<double> b(0.0);
    b.t; // compile error with version 1, fine with version 2
    return os;
}

int main() {
    A<int> a(0);
    cout << a << endl;
}
                        
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