Non-deduced context like 'boost::mpl::identity::type' in standard library?

Question

Cosider the following example I digged up here on StackOverflow

  template<typename T, typename Pred> 
  T const & clamp ( T const& val, 
    typename boost::mpl::identity<T>::type const & lo, 
    typename boost::mpl::identity<T>::type const & hi, Pred p )
  {
//    assert ( !p ( hi, lo ));    // Can't assert p ( lo, hi ) b/c they might be equal
    return p ( val, lo ) ? lo : p ( hi, val ) ? hi : val;
  }

where typename boost::mpl::identity<T>::type prevents the compiler from deducing T based on the type of the second and the third argument. This comes very handy for me, but I cannot use the Boost Library (please do not give me a hard time on that, as it is already a hard time because of that).

The question is now is something equivallent in the standard library directly which I just cannot find?

if there's nothing in the standard, it wouldn't be difficult to implement it yourself: the [synopsis here](https://www.boost.org/doc/libs/1_53_0/libs/mpl/doc/refmanual/identity.html) is pretty much all there is to it — kmdreko, Sep 09 '18 at 00:00
@kmdreko Totally right. I was just wondering about it being in the library as it struck me to be a nice thing to have. — , Sep 09 '18 at 00:04

score 2 · Accepted Answer · answered Sep 09 '18 at 04:16

C++20 will have std::type_identity. But you don't really have to wait for the standard library to have it. Its entire implementation is:

template< class T >
struct type_identity {
    using type = T;
};

template< class T >
using type_identity_t = typename type_identity<T>::type;

score 1 · Answer 2 · answered Sep 09 '18 at 00:41

1

boost::mpl::identity is a fairly straight-forward template that only provides a type identical to the provided template parameter.

It can be implemented as follows:

template <typename X>
struct identity
{
    typedef X type;
};

answered Sep 09 '18 at 00:41

kmdreko

42,554
6
57
106

score -1 · Answer 3 · answered Sep 09 '18 at 00:00

-1

std::common_type_t<T> works. From cppreference:

If sizeof...(T) is one (i.e., T... contains only one type T0), the member type names the same type as std::common_type<T0, T0>::type if it exists; otherwise there is no member type.

Thus, std::common_type_t will work for this

answered Sep 09 '18 at 00:00

Justin

24,288
12
92
142

1

Not that I'd necessarily recommend it. It's not obvious that it does this – Justin Sep 09 '18 at 00:02

Non-deduced context like 'boost::mpl::identity::type' in standard library?

3 Answers3