Simpler way to set multiple array slots to one value

Question

I\'m coding in C++, and I have the following code:

int array[30];
array[9] = 1;
array[5] = 1;
array[14] = 1;

array[8] = 2;
array[15] = 2;
array[23] = 2;
array[1         


        

Answer 1

Use overload operator << .

#include 
#include 
#include 

// value and indexes wrapper
template< typename T,  std::size_t ... Ints> struct _s{ T value; };

//deduced value type
template< std::size_t ... Ints,  typename T>
constexpr inline   _s  _ ( T const& v )noexcept { return {v}; }


// stored array reference
template< typename T, std::size_t N>
struct _ref
{
    using array_ref = T (&)[N];

    array_ref ref;
};


//join _s and _ref with << operator.
template< 
        template< typename , std::size_t ... > class IC, 
        typename U, std::size_t N, std::size_t ... indexes
        >
constexpr _ref operator << (_ref r, IC ic ) noexcept
{
    using list = bool[];
    return (  (void)list{ false, (  (void)(r.ref[indexes] = ic.value), false) ... }) , r ;

    //return r;
}


//helper function, for creating _ref from array.
template< typename T, std::size_t N>
constexpr inline _ref _i(T (&array)[N] ) noexcept { return {array}; }



int main()
{

   int a[15] = {0};

   _i(a) << _<0,3,4,5>(7) << _<8,9, 14>( 6 ) ;


   for(auto x : a)std::cout << x << "  " ;  
   //       0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
  //result: 7  0  0  7  7  7  0  0  6  6  0  0  0  0  6


  double b[101]{0};

  _i(b) << _<0,10,20,30,40,50,60,70,80,90>(3.14) 
        << _<11,21,22,23,24,25>(2.71) 
        << _<5,15,25,45,95>(1.414) ;
}