Question

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

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功能

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

Main.cpp的

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

归功于默认参数，您的专业化是

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

所以下面都要调用专业化

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

但非匹配类型会调用泛型：

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

具有专用成员函数的默认模板参数

1 个答案: