templated function pointer
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 MyC开发者_开发技巧lass::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>
.
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