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<typename T>
    auto operator()(const T& x) const -> decltype(-x) { return -x; }
};

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

struct complement
{
    template<typename T>
    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());

Answer 2

No (well, not directly). You need to use an adaptor, either old std::mem_fun (together with bind1st, IIRC) or std::bind/boost::bind.

std::transform(
    xs.begin(), xs.end(), ys.begin(),
    std::bind(&Class::member, &class_instance)
);

Answer 3

It is pretty easy to do if you wrap the call in lambda:

#include <algorithm>
#include <vector>

class C {
public:
  int op() const { return 1; }
};

class D {
  int op() { return 1; }
  void f() {
    std::vector<C> xs;
    std::vector<int> ys;
    std::transform(xs.begin(), xs.end(), ys.begin(),
      [](const C& x) { return x.op(); });
    std::transform(xs.begin(), xs.end(), ys.begin(),
      [this](const C& x) { return this->op(); });
  }
};