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, Mar 03 '15 at 14:16
Be aware that MSVC2013 has issues with the best practice answers to this problem in my experience. — Yakk - Adam Nevraumont, Mar 03 '15 at 14:47

score 10 · Accepted Answer · answered Mar 03 '15 at 14:01

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 03 '15 at 14:01

Angew is no longer proud of SO

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The 2nd template argument doesn't really need a name. But it is good – BЈовић Mar 03 '15 at 14:03
2

@BЈовић I know, but I generally use precisely this name as a bit of inline documentation. – Angew is no longer proud of SO Mar 03 '15 at 14:05
In conjunction with the note `14.8.1/7` of the standard, I find your answer very good. Thanks. – Raul Alonso Mar 03 '15 at 16:05

Best way to write a conditionally selectable constructor

1 Answers1