CODEWITHSUNDEEP
c++
EnDelt64
EnDelt64

Reputation: 1360

Conditions for template specialization type?

I've made a specialized template base on original template class like this:

Base template:

template <typename T>
class MyData {
public:
    MyData(T param) : data(param) {}

    T GetData() const {
        return data;
    }

private:
    T data;
};

Specialized template:

template <>
class MyData<char*> {
public:
    MyData(const char* param) {
        int len = static_cast<int>(strlen(param));
        data = new char[len + 1];
        strcpy_s(data, static_cast<rsize_t>(len + 1), param);
    }

    ~MyData() {
        delete[] data;
    }

    const char* GetData() const {
        return data;
    }

private:
    char* data;
};

This code works well, but I'm wondering if there any explicit conditions for the type of template specialization. For example, if I make function specialization, return type and all parameters' type must be the same.

template <>
const char* Add(const char* pLeft, const char* pRight) {
-----------     -----------        -----------
     |                |                  |
     +----------------+------------------+
    ...
}

In class MyData, template type of class is char* but the parameter type of constructor is const char*. However, there is no error while compiling the code. So I want to know is there any type conditions like this in class template specialization.

Upvotes: 1

Views: 58

Answers (1)

Dmitry Gordon
Dmitry Gordon

Reputation: 2324

There is a difference here between function specializations and class specializations. Class specializations may be completely different types with another set of members, functions and base classes. The following compiles well:

template <typename T>
class Foo
{
public:
    void Bar() {}

private:
    int val;
};

class Parent {};

template <>
class Foo<char> : public Parent
{
public:
    void BarBar() {}

private:
    double val;
};

int main()
{
    Foo<int> f;
    Foo<char> ff;
}

Upvotes: 1

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