Goal-Oriented Metaprogramming

So you want to...

C++ types are not first-class citizens. You cannot pass them as arguments or return them from functions, nor assign them to variables. This makes it challenging to manipulate them in a straightforward way as you would any other value.

C++ is such a large language (especially since C++ 11), that it's possible to do nearly anything. The trick is to figure out what language feature to use to accomplish your objective.

...write a function of types that returns a type

Since types are not first-class citizens in C++, this is impossible. Instead, define a template struct. The template parameters are the "function" arguments, and a type alias is the return type:

template <typename T, typename U>
struct make_a_tuple {
  using result = std::tuple<T, U>;

make_a_tuple is a "function" that takes two types and yields an std::tuple of those types:

auto tup = typename make_a_tuple<int, float>::result;

(since make_a_tuple::result is a dependent type, you need the typename keywork to tell C++ that make_a_tuple<...>::result is a type).

... write a function of types that returns a value

You can't write a function that takes a type as an argument and returns a value. There are two options here. If the value is tightly associated with the type, you can use a static constexpr member function, a static const member, or an enum (for integral types). I'm not sure why you would prefer one over the other.

template <unsigned I>
struct Int {
    static consteval int twice_function() {return I*2;}
    static const int twice_member = I * 2;
    enum : unsigned {twice_enum = I*2};

void foo() {

The second option is to define a template function that returns a value:

template <typename T, typename U>
consteval size_t size_of_pair() {
  return sizeof(std::make_pair<T(), U()>);

void foo() {
  size_t sz = size_of_pair<int, float>();

The second option is more attractive if you need to do some more complicated operation on types.

... operate conditionally on tupes

The simplest options is to use std::conditional from <type_traits>.

template <typename T, typename U>
struct larger_type {
  using type = std::conditional<
    sizeof(T) >= sizeof(U), // predicate (convertible to bool)
    T,  // if true
    U   // if false

void foo() {
  using Int = typename larger_type<long, int>::type;

This approach breaks down when anything you want to do in your conditional operation is not supported by both types. For example, using larger_type on void (can't use sizeof on void). The other more subtle failure is when your true or false type is invalid for some reason, regardless of whether it is selected by std::conditional.

... operate conditionally on types but std::conditional doesn't work

You have to use SFINAE.