Best way to write a conditionally selectable constructor

Question

I have the following code in C++:

struct A; struct B { B(){} template<typename T> B(T param){} };

I want the constructor template to be valid only when the typename T is convertible to the type A. What is the best way to accomplish this?

Sorry,I mixed C# and C++, I am working in both languages and I got confused. I just edited the question in order to fix the code. — Raul Alonso
– Raul Alonso, Commented Mar 3, 2015 at 14:16
Be aware that MSVC2013 has issues with the best practice answers to this problem in my experience. — Yakk - Adam Nevraumont
– Yakk - Adam Nevraumont, Commented Mar 3, 2015 at 14:47

Angew is no longer proud of SO · Accepted Answer · 2015-03-03 14:01:40Z

10

You want to enable the constructor if T is convertible to A? Use std::enable_if and std::is_convertible:

template < class T, class Sfinae = typename std::enable_if<std::is_convertible<T, A>::value>::type > B(T param) {}

This works by applying SFINAE; if T is not convertible to A, the substitution will fail and the constructor will be removed from the set of candidate overloads.

answered Mar 3, 2015 at 14:01

Angew is no longer proud of SO

172k18 gold badges377 silver badges478 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

BЈовић Over a year ago

The 2nd template argument doesn't really need a name. But it is good

Angew is no longer proud of SO Over a year ago

@BЈовић I know, but I generally use precisely this name as a bit of inline documentation.

Raul Alonso Over a year ago

In conjunction with the note 14.8.1/7 of the standard, I find your answer very good. Thanks.

Collectives™ on Stack Overflow

Best way to write a conditionally selectable constructor

1 Answer 1

3 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Linked

Related