c $ c> friend 作为函数模板如果你想匹配你定义的：

You have to declare friend as a function template if you want to match the one you defined:

template <typename U> // use U, so it doesn't *** with T
friend ostream& operator<<(ostream& os, MyVector<U> vt);

如果 friend 为类模板声明，不会使其成为函数模板。

If a friend function is declared for a class template, that does not make it a function template.