Pairwise bool and in c++ template

Question

I am writing a template that takes a arbitrary number of arguments and find the Boolean AND on these value.

template <bool... Vs> struct meta_bool_and;

template <bool V> struct meta_bool_and : std::integral_constant<bool, V> {}; 

template <bool V, bool... Vs> 
struct meta_bool_and : std::integral_constant<bool, V && meta_bool_and<Vs...>::value> {};

However, I failed to compile by the following message

 error: redeclared with 2 template parameters
 struct meta_bool_and : std::integral_constant<bool, V && meta_bool_and<Vs...>::value> {};

How can I fix this problem?

Jarod42 · Answer 1 · 2021-03-01T18:54:18.763

8

As alternatives, you may write it:

template <bool ... Bs>
using meta_bool_and = std::is_same<std::integer_sequence<bool, true, Bs...>,
                                   std::integer_sequence<bool, Bs..., true>>;

or in c++17:

template <bool ... Bs>
using meta_bool_and = std::integral_constant<bool, (Bs && ...)>;

edited Mar 01 '21 at 18:54

answered May 10 '16 at 20:48

Jarod42

203,559
14
181
302

The first one is very smart. Almost too smart for me… – 5gon12eder May 11 '16 at 00:39
`std::integral_sequence` ? Do you mean `std::integer_sequence` ? – 463035818_is_not_an_ai Mar 01 '21 at 17:13
1

@largest_prime_is_463035818:typo fixed, thanks. – Jarod42 Mar 01 '21 at 18:54

score 4 · Accepted Answer · answered May 10 '16 at 20:22

You have written a re-definition instead of a partial specialization. In order to provide a specialization, you must specify what properties you specialize on.

This will work:

#include <type_traits>

template <bool... Vs> struct meta_bool_and;

template <bool V> struct meta_bool_and<V> : std::integral_constant<bool, V> {};
//                                    ^^^

template <bool V, bool... Vs> 
struct meta_bool_and<V, Vs...> : std::integral_constant<bool, V && meta_bool_and<Vs...>::value> {}; 
//                  ^^^^^^^^^^

As an improvement, think whether you want to support the empty conjunction (typically defined as true). If so, don't specialize on meta_bool_and<bool> but on meta_bool_and<> (derived from std::true_type).

score 1 · Answer 3 · answered May 10 '16 at 20:22

Since these are specializations, they need to be declared as such. You can also make one the base case

template <bool V, bool... Vs>
struct meta_bool_and : std::integral_constant<bool, V && meta_bool_and<Vs...>::value> {};
// made base case

template <bool V>
struct meta_bool_and<V> : std::integral_constant<bool, V> {};
// specialization   ^^^

Pairwise bool and in c++ template

3 Answers3

Linked