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Is sizeof... allowed in template arguments for specialization?

I'm trying to do something along the lines of this using GCC 4.7 snapshot:

template <int n, int... xs>
struct foo { 
  static const int value = 0;
};

// partial specialization where n is number of ints in xs:

template <int... xs>
struct foo<sizeof...(xs), xs...> { // error: template argument ‘sizeof (xs ...)’
                                   //  involves template parameter(s)
  static const int value = 1;
};

template <int... xs>
struct foo<sizeof(xs), xs...> { // This compiles fine. sizeof(xs) is sizeof int 
                                // even though packs aren't expanded
  static const int value = 2;
};

The error is strange because sizeof instead of sizeof... works in this case. Both seem like they could be easily computed during compile time.

Is the compiler correct that I can't use sizeof... in template arguments for specialization?

like image 214
Pubby Avatar asked Oct 23 '22 18:10

Pubby


1 Answers

I'm going to assume this is a compiler issue after reading this post.

A partially specialized non-type argument expression shall not involve a template parameter of the partial specialization except when the argument expression is a simple identifier.

Which is being disputed here.

GCC is either incorrectly unpacking the parameter pack, or evaluating sizeof prematurely.

Response to bug report I filed may be helpful.

like image 51
Pubby Avatar answered Oct 27 '22 00:10

Pubby