构造函数模板专门化对析构函数给出了未定义的引用

Question

当我不专门化构造函数(而是其他方法)时，一切都很好:

Test.h:

struct Type_1 {};
struct Type_2 {};
template<typename Type>
class Test {
public:
    Test();
    ~Test();
    void print();
}

Test.cpp:

template<typename Type>
Test<Type>::Test() {}
template<typename Type>
Test<Type>::~Test() {}
template<>
Test<Type_1>::print(){ // Type_1 specific print }
template<>
Test<Type_2>::print(){ // Type_2 specific print }

工作得很好，只要模板是Type_1或Type_2，他就创建类并相应地调用print()。

但是如果我尝试专化构造函数，他说"对析构函数的未定义引用"，但是析构函数应该自动推断参数类型，对吗?

Test.cpp改变:

template<>
Test<Type_1>::Test(){ // constructor when Type_1 }
template<>
Test<Type_2>::Test(){ // constructor when Type_2 }
template<typename Type>
Test<Type>::~Test(){ // should work for all right? }

我的问题是:如果我模板专门化一个构造函数，我必须专门化的析构函数，以及，即使它是相同的?还有别的办法吗?

Answer 1

不需要定义任何成员函数的专门化版本。你可以这样做:

template<typename Type>
class Test {
public:
    Test();
    ~Test() 
    {
        std::cout << "Default Dtor called." << std::endl;
    }
    void print();
};
//! These should go either in the class declaration or in a separate ipp/inl file.
//! You can put these in a .cpp if you use extern templates, but they might not be
//! supported on your compiler.
//! NOTE: If you use ipp/inl or declare these in the header but still outside the
//! class declaration, be sure to use inline on the member function definitions.
struct Type_1{};
struct Type_2{};
template <>
inline Test<Type_1>::Test()
{
    std::cout << "Type_1 Ctor called" << std::endl;
}
template <>
inline Test<Type_1>::~Test()
{
    std::cout << "Type_1 Dtor called" << std::endl;
}
template <>
inline Test<Type_2>::Test()
{
    std::cout << "Type_2 Ctor called" << std::endl;
}
int main()
{
    {
        Test<Type_1> t1;
    }
    {
        Test<Type_2> t2;
    }
}

输出:

Type_1 Ctor调用

Type_1医生呼叫

Type_2函数调用

Default doctor called.

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