Default function that just returns the passed value?

Question

As a lazy developer, I like to use this trick to specify a default function:

template <class Type, unsigned int Size, class Function = std::less<Type> >
void arrange(std::array<Type, Size> &x, Function&& f = Function())
{
    std::sort(std::begin(x), std::end(x), f);
}

But I have a problem in a very particular case, which is the following:

template <class Type, unsigned int Size, class Function = /*SOMETHING 1*/>
void index(std::array<Type, Size> &x, Function&& f = /*SOMETHING 2*/)
{
    for (unsigned int i = 0; i < Size; ++i) {
        x[i] = f(i);
    }
}

In this case, I would like the default function to be the equivalent of: [](const unsigned int i){return i;} (a function that just returns the passed value).

In order to do that, what do I have to write instead of /*SOMETHING 1*/ and /*SOMETHING 2*/?

Answer 1

This is called the identity function. Unfortunately, it is not part of the C++ standard, but you can easily build one yourself.

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If you happen to use g++, you can activate its extensions with -std=gnu++11 and then

#include <array>
#include <ext/functional>

template <class Type, std::size_t Size, class Function = __gnu_cxx::identity<Type> >
void index(std::array<Type, Size> &x, Function&& f = Function())
{
    for (unsigned int i = 0; i < Size; ++i) {
        x[i] = f(i);
    }
}

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Maybe it will be available in C++20, see std::identity. Until then you may look at boost's version at boost::compute::identity.