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)在这个实例中没有任何意义。)
答案 0 :(得分:8)
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> {
};
答案 1 :(得分:2)
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
}