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

Question

template < class InputIterator, class OutputIterator, class UnaryOperator >
  OutputIterator transform ( InputIterator first1, InputIterator last1,
                             OutputIterator result, UnaryOperator op );

Can op be a member function? If so, how do I call it?

http://stackoverflow.com/questions/1762781/mem-fun-and-bind1st-problem — q0987, Jan 26 '12 at 20:06
You question is slightly ambiguous. What class are you hoping to make `op` a member of? The class of `*first1`, the class of `*result`, the class of `*this`, or some other class? — Robᵩ, Jan 26 '12 at 20:22
@q0987 - see my answer. It describes the case of `op` being a member of the class of `*first1` and the case of `op` being a member of `*this`. You should be able to generalize from there. — Robᵩ, Jan 26 '12 at 20:33

score 4 · Answer 1 · answered Jan 26 '12 at 19:59

4

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)
);

answered Jan 26 '12 at 19:59

Cat Plus Plus

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1

Isn't `class_instance` supposed to be supplied by `std::transform`? So `std::mem_fun` by itself would probably be more appropriate than `bind1st`. – Ben Voigt Jan 26 '12 at 20:18
Replace `&class_instance` with `std::placeholders::_1`. – Xeo Jan 26 '12 at 20:19
@Ben: `mem_fn` is the appropriate Boost/C++11 version. – Xeo Jan 26 '12 at 20:22
@BenVoigt: No idea, OP can be about either, really. – Cat Plus Plus Jan 26 '12 at 20:27
1

IIRC C++11 deprecated `std::mem_fun()`. – wilhelmtell Jan 27 '12 at 19:06
I really envy all you folks here on StackOverflow who're able to use the fancy new C++11 features in their projects, are you all students or hobbyists? If you can use this stuff in your daytime job, please let me know where you work. ;-) – Frerich Raabe Jan 27 '12 at 19:12

score 3 · Answer 2 · answered Jan 26 '12 at 20:18

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(); });
  }
};

Ben Voigt · Answer 3 · 2012-01-26T20:13:03.243

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());

@q0987: I showed examples of lambdas. Was something wrong with them? — Ben Voigt, Jan 30 '12 at 03:34

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

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