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

You could use boost::int_t:

using coord_squared_t = boost::int_t<sizeof(coord_t)*CHAR_BIT*2>::least;

Answer 2

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

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

template <> struct next_size<int_least8_t>  : tag<int_least16_t> { };
template <> struct next_size<int_least16_t> : tag<int_least32_t> { };
template <> struct next_size<int_least32_t> : tag<int_least64_t> { };
template <> struct next_size<int_least64_t> : tag<???> { };

// + others if you want the other int types

And then:

using coord_squared_t = next_size_t<coord_t>;

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Alternatively you can specialize based on number of bits:

template <size_t N> struct by_size : by_size<N+1> { };
template <size_t N> using by_size_t = typename by_size<N>::type;
template <class T> struct tag { using type = T; };

template <> struct by_size<8>  : tag<int_least8_t> { };
template <> struct by_size<16> : tag<int_least16_t> { };
template <> struct by_size<32> : tag<int_least32_t> { };
template <> struct by_size<64> : tag<int_least64_t> { };

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)>;