How can I create a constexpr function that returns a type (to be used in a template parameter)

Question

I'm looking for some way to create a class, with a template parameter type, based on a template parameter number.

What I'm trying to do is something like this:

template<size_t n>
constexpr auto type_from_size() {
    if(n < 256) {
        return uint8_t;
    } else {
        return uint16_t;
    }
}

template<size_t n>
class X {
    type_from_size<n>() t;
}

X<500> x;
x.t = 500;

So, in the code above, the constexpr function type_from_size() would receive the number 500 and would return the type uint16_t, and this would be the type of the member X.t.

I know this is obviously terrible code, but is this possible using templates?

Answer 1

constexpr function cannot return a type directly but a simple wrapper will work.

template <typename T>
struct TypeWrapper {
    using type = T;
};

template <size_t n>
constexpr auto type_from_size_func() {
    if constexpr (n <= 0xff) {
        return TypeWrapper<uint8_t>{};
    } else {
        if constexpr (n <= 0xffff) {
            return TypeWrapper<uint16_t>{};
        } else {
            return TypeWrapper<uint32_t>{};
        }
    }
}

template <size_t N>
using type_from_size = typename decltype(type_from_size_func<N>())::type;

usage

type_from_size<123> x; /*uint8_t*/
type_from_size<1234> y; /*uint16_t*/
type_from_size<12345678> z; /*uint32_t*/

Answer 2

Let's just go overkill. Start with a chooser:

template <int I> struct choice : choice<I + 1> { };
template <> struct choice<10> { };

struct otherwise { otherwise(...) { } };

Then create a cascading series of overloads that return a type. The choice ensures that the smallest type will be selected first, without having to write two-sided ranges for all the intermediate integer types:

template <class T> struct tag { using type = T; }
template <size_t N> using size_t_ = std::integral_constant<size_t, N>;

template <size_t N, class = std::enable_if_t<(N < (1ULL << 8))>>
constexpr tag<uint8_t> tag_from_size(size_t_<N>, choice<0> ) { return {}; }

template <size_t N, class = std::enable_if_t<(N < (1ULL << 16))>>
constexpr tag<uint16_t> tag_from_size(size_t_<N>, choice<1> ) { return {}; 

template <size_t N, class = std::enable_if_t<(N < (1ULL << 32))>>
constexpr tag<uint32_t> tag_from_size(size_t_<N>, choice<2> ) { return {}; }

template <size_t N>
constexpr tag<uint64_t> tag_from_size(size_t_<N>, otherwise) { return {}; }

And then you can write the top level one that dispatches:

template <size_t N>
using type_from_size_t = typename decltype(tag_from_size(size_t_<N>{}, choice<0>{}))::type;

And use it:

template <size_t N>
class X {
    type_from_size_t<N> t;
};

Answer 3

First write static_if<bool>( A, B ).

Next, write template<class T> struct tag_type{using type=T;}; and supporting code.

template<size_t n>
constexpr auto type_from_size() {
  return static_if< (n<256) >( 
    tag<std::uint8_t>,
    tag<std::uint16_t>
  );
}

now returns a different tag type based on the value of n.

To use:

template<size_t n>
class X {
  typename decltype(type_from_size<n>())::type t;
}

or write a quick alias:

template<size_t n> type_from_size_t = typename decltype(type_from_size<n>())::type;

Here is the code for tag_type and static_if:

template<class T>struct tag_type{using type=T;};
template<class T>constexpr tag_type<T> tag{};

template<bool b>using bool_t=std::integral_constant<bool, b>;

template<class T, class F>
constexpr T static_if( bool_t<true>, T t, F f ) {
  return t;
}
template<class T, class F>
constexpr F static_if( bool_t<false>, T t, F f ) {
  return f;
}
template<bool b, class T, class F>
constexpr auto static_if( T t, F f ) {
  return static_if( bool_t<b>, t, f );
}
template<bool b, class T>
constexpr auto static_if( T t ) {
  return static_if<b>( t, [](auto&&...){} );
}

and done.

Note we can also do static_case. :)

Answer 4

A function cannot return a type. You should use a template.

For a selection between only two types, the built-in std::conditional is sufficient.

#include <type_traits>
#include <cstdint>

template <size_t n>
using type_from_size = typename std::conditional<(n < 256), uint8_t, uint16_t>::type;
// ^ if `n < 256`, the ::type member will be typedef'ed to `uint8_t`.
//                 otherwise, it will alias to `uint16_t`.
//   we then give a convenient name to it with `using`.

template <size_t n>
struct X {
    type_from_size<n> t;
    // ^ use the template
};

---

If you need to support more than two values, you can change multiple conditional together like an if/else if/else chain, but OH MY EYES

template <size_t n>
using type_from_size =
    typename std::conditional<(n <= 0xff), uint8_t,
        typename std::conditional<(n <= 0xffff), uint16_t,
            typename std::conditional<(n <= 0xffffffff), uint32_t,
                uint64_t
            >::type
        >::type
    >::type;

---

You could also use specialization together with std::enable_if (SFINAE) to make it more "low-level":

template <size_t n, typename = void>
struct type_from_size_impl;
// Declare a "size_t -> type" function.
//  - the `size_t n` is the input
//  - the `typename = void` is a placeholder
//    allowing us to insert the `std::enable_if` condition.

template <size_t n>
struct type_from_size_impl<n, typename std::enable_if<(n <= 0xff)>::type> {
    using type = uint8_t;
};
// We add a partial specialization
//  - in `std::enable_if<c>::type`, if `c` is true, `::type` will be typedef'ed to `void`
//  - otherwise, `::type` will not be defined.
//  - if `::type` is not defined, substitution failed,
//    meaning we will not select this specialization

template <size_t n>
struct type_from_size_impl<n, typename std::enable_if<(n > 0xff && n <= 0xffff)>::type> {
    using type = uint16_t;
};

template <size_t n>
struct type_from_size_impl<n, typename std::enable_if<(n > 0xffff && n <= 0xffffffff)>::type> {
    using type = uint32_t;
};

template <size_t n>
struct type_from_size_impl<n, typename std::enable_if<(n > 0xffffffff)>::type> {
    using type = uint64_t;
};

template <size_t n>
using type_from_size = typename type_from_size_impl<n>::type;
// Here we want to find a specialization of `type_from_size_impl<n>`
// All 4 specializations will be tried.
// If only one specialization works, we will use that one
// (Which is why we need to ensure the ranges are not overlapping
//  otherwise the compiler will complain)
// Then we take the `::type` out the complete this "type-level function".

Answer 5

Definitely. Here's a more flexible way of doing it, you can add as many ranges as you wish as long as they don't overlap.

template <std::size_t N, class = void>
struct TypeForSize_;

template <std::size_t N>
struct TypeForSize_<N, std::enable_if_t<
    (N <= 255)
>> { using type = std::uint8_t; };

template <std::size_t N>
struct TypeForSize_<N, std::enable_if_t<
    (N > 255 && N <= 65535)
>> { using type = std::uint16_t; };

template <std::size_t N>
using TypeForSize = typename TypeForSize_<N>::type;

Using a size for which no type has been defined will result in a compile-time error.