6

假设我有以下课程:

template <typename T>
class MyClass
{
public:
    void SetValue(const T &value) { m_value = value; }

private:
    T m_value;
};

如何为 T=float (或任何其他类型)编写函数的专用版本?

注意:一个简单的重载是不够的,因为我只希望函数可用于 T=float(即 MyClass::SetValue(float) 在这种情况下没有任何意义)。

4

2 回答 2

9
template <typename T>
class MyClass {
private:
    T m_value;

private:
    template<typename U>
    void doSetValue (const U & value) {
        std::cout << "template called" << std::endl;
        m_value = value;
    }

    void doSetValue (float value) {
        std::cout << "float called" << std::endl;
    }

public:
    void SetValue(const T &value) { doSetValue (value); }

};

或(部分模板专业化):

template <typename T>
class MyClass
{
private:
    T m_value;

public:
    void SetValue(const T &value);

};

template<typename T>
void MyClass<T>::SetValue (const T & value) {
    std::cout << "template called" << std::endl;
    m_value = value;
}

template<>
void MyClass<float>::SetValue (const float & value) {
    std::cout << "float called" << std::endl;
}

或者,如果您希望函数具有不同的签名

template<typename T>
class Helper {
protected:
    T m_value;
    ~Helper () { }

public:
    void SetValue(const T &value) {
        std::cout << "template called" << std::endl;
        m_value = value;
    }
};

template<>
class Helper<float> {
protected:
    float m_value;
    ~Helper () { }

public:
    void SetValue(float value) {
        std::cout << "float called" << std::endl;
    }
};

template <typename T>
class MyClass : public Helper<T> {
};
于 2012-10-14T08:14:52.550 回答
3

你当然可以。只是它应该是一个过载:)

template <typename T>
class MyClass
{
public:
    template<class U> 
    void SetValue(const U &value) { m_value = value; }
    void SetValue(float value) {do special stuff} 
private:
    T m_value;
};

int main() 
{
     MyClass<int> mc;
     mc.SetValue(3.4); // the template member with U=double will be called
     mc.SetValue(3.4f); // the special function that takes float will be called

     MyClass<float> mcf; //compiles OK. If the SetValue were not a template, 
     // this would result in redefinition (because the two SetValue functions
     // would be the same   
}
于 2012-10-13T22:11:44.380 回答