I am writing a template class and want to find out whether template argument is default constructible is there any way to do it ?
The code is something like following
template <class C>
class A
{
createObj()
{
C* objPtr = NULL;
// If default constructible then create object else let it remain NULL
}
};
Update: I have tried using code given in this question but it doesn't work, to be precise if return default constructible even for those classes which aren't, I have no idea why is that happening.
This is a classical case for SFINAE and enable_if
.
In another answer, Potatoswatter has already posted a type trait is_default_constructible
that can be reused here:
void createObj(
typename enable_if_c<is_default_constructible<C>::value, void>::type* dummy = 0)
{
C* objPtr = new C();
}
void createObj(
typename disable_if_c<is_default_constructible<C>::value, void>::type* dummy = 0)
{
C* objPtr = 0;
}
Or, if your function has a non-void return type T
(thanks to DeadMG) you can omit the dummy default argument:
typename enable_if_c<is_default_constructible<C>::value, T>::type createObj()
{
C* objPtr = new C();
}
typename disable_if_c<is_default_constructible<C>::value, T>::type createObj()
{
C* objPtr = 0;
}
SFINAE means that a template which cannot be instantiated for a given type, will not. enable_if_c
basically results in a valid type if and only if its argument is true
. Now we use the metafunction is_default_constructible
to test whether the type C
has a default constructor. If so, enable_if_c<…>::type
will result in a valid type, otherwise it will not.
The C++ compiler will thus only see one of the two functions, to wit the one that is usable in your context. See the documentation of enable_if
for more details.
This is a case of SFINAE, you use a function overload, one that takes ... as a parameter, one that calls the method you are looking for.
In any case this particular question is answered `here
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