C ++中没有模板typedef。在C ++ 0x中有这样的扩展。同时，执行

 模板< typename T> 
 struct TFunc 
 {
 typedef void（MyClass :: * type）（T param）; 
}; 
 

并使用 TFunc< T> :: type 每当您使用 TFunc< T> 时，在之前加上 typename

I have an approach to call delayed function for class:

//in MyClass declaration:
typedef void (MyClass::*IntFunc) (int value);
void DelayedFunction (IntFunc func, int value, float time);
class TFunctorInt
{
public:
    TFunctorInt (MyClass* o, IntFunc f, int v) : obj (o), func (f), value (v) {}
    virtual void operator()();
protected:
    MyClass* obj;
    IntFunc func;
    int value;
};
//in MyClass.cpp file:
void MyClass::DelayedFunction (IntFunc func, int value, float time)
{
    TFunctorBase* functor = new TFunctorInt (this, func, value);
    DelayedFunctions.push_back (TDelayedFunction (functor, time)); // will be called in future
}
void MyClass::TFunctorInt::operator()()
{
    ((*obj).*(func)) (value);
}

I want to make templated functor. And the first problem is that:

template <typename T>
typedef void (MyClass::*TFunc<T>) (T param);

Causes compiler error: "template declaration of 'typedef'". What may be a solution?

PS: The code based on http://www.coffeedev.net/c++-faq-lite/en/pointers-to-members.html#faq-33.5

There are no template typedefs in C++. There is such an extension in C++0x. In the meantime, do

template <typename T>
struct TFunc
{
    typedef void (MyClass::*type)(T param);
};

and use TFunc<T>::type (prefixed with typename if in a dependant context) whenever you would have used TFunc<T>.