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

Just for the fun of it I created a somewhat different approach which needs a bit of infrastructure allowing initialization like so:

double array[40] = {};
"9 5 14"_idx(array) = 1;
"8 15 23 12"_idx(array) = 2;

If the digits need to be separated by commas, there is a small change needed. In any case, here is the complete code:

#include 
#include 
#include 
#include 

template 
class assign
{
    int  d_indices[Size];
    int* d_end;
    T*   d_array;
    void operator=(assign const&) = delete;
public:
    assign(char const* base, std::size_t n)
        : d_end(std::copy(std::istream_iterator(
                      std::istringstream(std::string(base, n)) >> std::skipws),
                          std::istream_iterator(), this->d_indices))
        , d_array()
    {
    }
    assign(assign* as, T* a)
        : d_end(std::copy(as->begin(), as->end(), this->d_indices))
        , d_array(a) {
    }
    assign(assign const& o)
        : d_end(std::copy(o.begin(), o.end(), this->d_indices))
        , d_array(o.d_array)
    {
    }
    int const* begin() const { return this->d_indices; }
    int const* end() const   { return this->d_end; }
    template 
    assign operator()(A* array) {
        return assign(this, array);
    }
    void operator=(T const& value) {
        for (auto it(this->begin()), end(this->end()); it != end; ++it) {
            d_array[*it] = value;
        }
    }
};

assign<30> operator""_idx(char const* base, std::size_t n)
{
    return assign<30>(base, n);
}

int main()
{
    double array[40] = {};
    "1 3 5"_idx(array) = 17;
    "4 18 7"_idx(array) = 19;
    std::copy(std::begin(array), std::end(array),
              std::ostream_iterator(std::cout, " "));
    std::cout << "\n";
}