Why can't I use a constexpr data member pointer as a template argument in C++11?

Question

Please consider the following code:

template <typename T, typename P, T P:: *s> struct H {};

struct AA { int i; };

int main()
{
  typedef int AA::*PI;
  constexpr PI pi = &AA::i;

  H<int, AA, &AA::i> h1;    // OK
  // H<int, AA, pi> h2;     // compile error
}

I have member pointer pi pointing to AA::i. pi is a constexpr variable. Why can't I use it as a template parameter, even though using &AA::i directly works?

Mike Seymour · Accepted Answer · 2015-04-10 08:54:15Z

7

Because those are the rules, at least in C++11; 14.3.2/1 only allows "a pointer to member expressed as described in 5.3.1", which describes the &AA::i syntax.

This has changed in the latest draft, and now the requirement for any type is just "a converted constant expression of the type of the template-parameter", under which your code would be fine.

I don't know whether or not this change is in C++14, since I don't yet have access to that published standard.

edited Apr 10, 2015 at 8:54

answered Apr 10, 2015 at 8:48

Mike Seymour

256k30 gold badges467 silver badges658 bronze badges

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

3 Comments

Kerrek SB Over a year ago

The DIS for C++14 is N4141 I think, which you can check out from GitHub yourself.

Kerrek SB Over a year ago

That said, I compared the two and indeed the change was only applied after C++14.

Zhipeng YANG Over a year ago

Thanks for your sincere answer!

Collectives™ on Stack Overflow

Why can't I use a constexpr data member pointer as a template argument in C++11?

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related