CODEWITHSUNDEEP
c++c++11stdfunctor
Bernard
Bernard

Reputation: 5666

C++ functor that returns a reference to its input parameter

Consider this function, which is supposed to multiply each number in a range by a supplied value, and optionally take in a functor to get a reference to the number to be multiplied:

template <typename TIter, typename TNumber, typename TTransformer = self_reference<iterator_traits<TIter>::value_type>>
void multiply_range(TIter begin, TIter end, TNumber multiplicand, TTransformer transformer = TTransformer()) {
    for_each(begin, end, [multiplicand, &transformer](typename iterator_traits<TIter>::value_type val){
        transformer(val) *= multiplicand;
    }
}

Is there something in the standard library that does what self_reference is trying to do, i.e. a functor that returns a reference to its input parameter?

(The standard library has things like std::less and std::plus, but am I not asking for something even more basic?)

So, the following code should work:

int x[4] = {1, 3, 5, 7};
multiply_range(x, x + 4, 2);
// x will become {2, 6, 10, 14}

vector<int> y[2] = {{1, 2}, {3, 4, 5}};
multiply_range(y, y + 2, 2, [](vector<int>& v) -> int& { return v[0]; });
// y will become {{2, 2}, {6, 4, 5}}

C++11 and C++14 are fine too.

Upvotes: 3

Views: 184

Answers (1)

Cubic
Cubic

Reputation: 15673

a functor that returns a reference to its input parameter

is quite different from a

C++ Functor that Returns Itself

What you're looking for is most commonly known as the "identity" transform(er, or functor), and is not part of the C++ standard library.

Upvotes: 4

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