c++ - 具有专门成员函数的默认模板参数

Question

我有一个类模板 stack，带有默认模板参数和 pop() 函数，它是专门的成员功能

template <typename T,typename CONT = vector<T> >
class Stack
{
public:
 void push(T arg);
 T top()const;
 void pop();
 bool isEmpty() const;
private:
 CONT elems_;
};

template <typename T,typename CONT>
void Stack<T,CONT>::pop()
{
 elems_.pop_back();
}

//专门的pop函数

template<>
void Stack<int>::pop()
{
 cout << "Called Specialized ";
 elems_.pop_back();
}

main.cpp

Stack<int> mystack;
mystack.push(10);
mystack.pop(); ---> this calls specialized one Why ?

Stack<int,vector<int>> mystack;
mystack.push(10);
mystack.pop(); ---> this calls  template one Why ?

Answer 1

由于默认参数，您的专长是

template<>
void Stack<int, std::vector<int>>::pop()
{
    cout << "Called Specialized ";
    elems_.pop_back();
}

所以以下都称为特化

Stack<int> mystack1; // Stack<int, std::vector<int>>
mystack1.push(10);
mystack1.pop(); // ---> this calls specialized

Stack<int, vector<int>> mystack2;
mystack2.push(10);
mystack2.pop(); // ---> this calls specialized

但是不匹配的类型会调用泛型:

Stack<int, list<int>> mystack3;
mystack3.push(10);
mystack3.pop(); // ---> this calls generic one

Stack<char> mystack4; // Stack<char, std::vector<char>>
mystack4.push(10);
mystack4.pop(); // ---> this calls generic one