C++ integer type twice the width of a given type

Question

In this example, coord_squared_t is the alias for an integer type with at least twice the size of the integer type coord_t:

typedef         


        

Answer 1

If you don't want to use Boost, you could just implement this manually with some specializations:

template  struct next_size;
template  using next_size_t = typename next_size::type;
template  struct tag { using type = T; };

template <> struct next_size  : tag { };
template <> struct next_size : tag { };
template <> struct next_size : tag { };
template <> struct next_size : tag { };

// + others if you want the other int types

And then:

using coord_squared_t = next_size_t;

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Alternatively you can specialize based on number of bits:

template  struct by_size : by_size { };
template  using by_size_t = typename by_size::type;
template  struct tag { using type = T; };

template <> struct by_size<8>  : tag { };
template <> struct by_size<16> : tag { };
template <> struct by_size<32> : tag { };
template <> struct by_size<64> : tag { };

This way, something like by_size<45>::type is int_least64_t due to inheritance. And then this becomes just like the Boost answer:

using coord_squared_t = by_size_t<2 * CHAR_BIT * sizeof(coord_t)>;