I've got problem with creating function that for given type, if it's derived from other one do something and for all other cases do something other. My code:
class BaseClass {};
class DerivedClass : public BaseClass {};
template <typename T>
void Function(typename std::enable_if<std::is_base_of<BaseClass, T>::value, T>::type && arg) {
std::cout << "Proper";
}
template <typename T>
void Function(T && arg) {
std::cout << "Improper";
}
void test() {
Function(DerivedClass{});
}
As mentioned in the comments to the question, SFINAE expressions won't work the way you did it.
It should be instead something like this:
template <typename T>
typename std::enable_if<std::is_base_of<BaseClass, T>::value>::type
Function(T && arg) {
std::cout << "Proper" << std::endl;
}
template <typename T>
typename std::enable_if<not std::is_base_of<BaseClass, T>::value>::type
Function(T && arg) {
std::cout << "Improper" << std::endl;
}
SFINAE expressions will enable or disable Function
depending on the fact that BaseClass
is base of T
. Return type is void
in both cases, for it's the default type for std::enable_it
if you don't define it.
See it on coliru.
Other valid alternatives exist and some of them have been mentioned in other answers.