Can UnaryOperator be a member function when std::transform is called

Question

Based on std::transform

template < class InputIterator, class OutputIterator, class UnaryOperator >
  OutputIterator transform ( InputIterator first1, Inpu         


        

Answer 1

You need a helper object, like std::less but for a unary operator.

C++11 lambdas make this incredibly easy:

std::transform(xs.begin(), xs.end(), ys.begin(), [](the_type x){ return -x; });
std::transform(xs.begin(), xs.end(), ys.begin(), [](the_type x){ return !x; });
std::transform(xs.begin(), xs.end(), ys.begin(), [](the_type x){ return ~x; });

Or, use these flexible helpers:

struct negate
{
    template
    auto operator()(const T& x) const -> decltype(-x) { return -x; }
};

struct invert
{
    template
    auto operator()(const T& x) const -> decltype(!x) { return !x; }
};

struct complement
{
    template
    auto operator()(const T& x) const -> decltype(~x) { return ~x; }
};

std::transform(xs.begin(), xs.end(), ys.begin(), negate());
std::transform(xs.begin(), xs.end(), ys.begin(), invert());
std::transform(xs.begin(), xs.end(), ys.begin(), complement());