C++17: integer_sequence usage with compilation error

Question

I wish to use a integer_sequence to judge if a range of numbers are all under a certain value: is_range() will return true, else return false, like below:

#include<utility> 
#include<iostream> 
using namespace std; 
template <std::size_t N, std::size_t... Ix> 
bool in_range(std::index_sequence<Ix...>) { 
   return ((Ix < N) && ...); 
} 
int main() 
{ 
     cout<<in_range<10>({1,2,30})<<endl; 
     cout<<in_range<10>(1,2,3)<<endl; 
     return 0; 
}

I used clang3.8 to compile it, but failed.

$ clang++ m.cpp -std=c++1z 
m.cpp:5:37: error: template argument for template type parameter must be a type 
bool in_range(std::integer_sequence<Ix...>) { 
                                     ^~~~~ 
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.1/../../../../include/c++/5.3.1/utility:229:21: note: 
       template parameter is declared here 
   template<typename _Tp, _Tp... _Idx> 
                     ^ 
m.cpp:10:11: error: no matching function for call to 'in_range' 
     cout<<in_range<10>({1,2,30})<<endl; 
           ^~~~~~~~~~~~ 
m.cpp:11:11: error: no matching function for call to 'in_range' 
     cout<<in_range<10>(1,2,3)<<endl; 
           ^~~~~~~~~~~~ 
3 errors generated.

How should I correct my code? I suppose my understanding of fold expression is incorrect

How to correct it?

The `std::index_sequence` expects a type as its first template argument. You are passing the `size_t` value `Ix`, which is an integer. — Marvin Sielenkemper, Jul 13 '16 at 04:05
@MarvinSielenkemper : `std::integer_sequence` expects a type as its first template argument – `std::index_sequence` is the former for `std::size_t`. — ildjarn, Jul 13 '16 at 21:05

Praetorian · Accepted Answer · 2016-07-13T05:05:52.840

There's no need for index_sequence here, you can just pass the list of numbers to be compared as template arguments.

template <std::size_t N, std::size_t... Ix> 
bool in_range() { 
   return ((Ix < N) && ...); 
}
cout<<in_range<10,1,2,30>()<<endl;

Or if you want to pass them as arguments to the function template

template <std::size_t N, typename... Ix> 
bool in_range(Ix... ix) { 
   return ((ix < N) && ...); 
}
cout<<in_range<10>(1U,2U,30U)<<endl;

Finally, if you want to be able to pass a braced-init-list to in_range you should accept an initializer_list<size_t>. Otherwise, template argument deduction will fail because a braced-init-list is not an expression, and so it has no type.

template <std::size_t N> 
constexpr bool in_range(std::initializer_list<std::size_t> ix) { 
    for(auto i : ix) {
        if(i >= N) return false;
    }
    return true;
}
cout<<in_range<10>({1,2,30})<<endl;

Live demo

C++17: integer_sequence usage with compilation error

1 Answers1