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How to check if a template argument is default constructible

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.

like image 347
Gaurav Avatar asked Oct 14 '22 18:10

Gaurav


2 Answers

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.

like image 199
Konrad Rudolph Avatar answered Oct 18 '22 03:10

Konrad Rudolph


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

like image 23
CashCow Avatar answered Oct 18 '22 03:10

CashCow