I have been trying to create this class which can either use the default functor as an argument or the user can provide one if he wants. But I am unable to pass function pointer as my template argument. Can you please help me in understanding what I am missing.
template <typename T>
struct CheckFunctor
{
bool operator()(T obj)
{
return true;
}
};
template <typename _Ty,
class _Pr = CheckFunctor<_Ty>
>
class MyClass
{
typedef _Ty mapped_type;
typedef _Pr CanBeCleaned_type;
_Ty data;
CanBeCleaned_type predicate;
public:
void SomeMethod()
{
if( predicate(data))
{
std::cout << "Do something";
}
}
MyClass(_Ty timeOutDuration, _Pr pred = _Pr())
: data( timeOutDuration), predicate( pred)
{}
};
template< typename T>
struct CheckEvenFunctor
{
bool operator()(T val)
{
return (val%2 == 0);
}
};
bool CheckEven( int val)
{
return (val%2 == 0);
}
int main()
{
//Usage -1
MyClass<int> obj1( 5);
//Usage- 2
MyClass< int, CheckEven> obj2(6, CheckEven); //Error: 'CheckEven' is not a valid template type argument for parameter '_Pr'
//Usage -3
MyClass<int, CheckEvenFunctor<int>>( 7);
}
You are trying to pass CheckEven as a type parameter even though CheckEven is not a type but a function (of type bool(int)). You should define the type as a pointer to the type of function that you are passing. decltype is handy here:
MyClass< int, decltype(&CheckEven)> obj2(6, CheckEven);
You can also create a factory function and let the compiler deduce the template parameters:
template<class T, class F>
MyClass<T, F> makeClass(T timeOutDuration, F pred) {
return {timeOutDuration, pred};
}
auto obj2 = makeClass(6, CheckEven);
Collected from the Internet
Please contact [email protected] to delete if infringement.
Comments