Pass std algos predicates by reference in C++

Question

I am trying to remove elements from a std::list and keep some stats of deleted elements.

In order to do so, I use the remove_if function from the list, and I have a predicate. I would like to use this predicate to gather statistics. Here is the code for the predicate:

  class TestPredicate
  {
  private:
    int limit_;

  public:
    int sum;
    int count;
    TestPredicate(int limit) : limit_(limit), sum(0), count(0) {}

    bool operator() (int value) 
    {
      if (value >= limit_)
      {
        sum += value;
        ++count;       // Part where I gather the stats
        return true;
      }
      else
        return false;
    }
  };

And here is the code for the algo:

std::list < int > container;
container.push_back(11);
TestPredicate pred(10);
container.remove_if(pred)
assert(pred.count == 1);

Unfortunately, the assertion is false because the predicate is passed by value. Is there a way to force it to be passed by reference ?

Answer 1

Pass a reference wrapper, available from <functional>:

container.remove_if(std::ref(pred));

---

If you only have C++98/03 but your compiler has TR1, you can use <tr1/functional> and std::tr1::ref if you make a small amendment to your predicate:

#include <tr1/functional>

class TestPredicate : public std::unary_function<int, bool>
{                 //^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  // ...
}

container.remove_if(std::tr1::ref(pred));

---

If all else fails, then you can hack up a manual solution with relative ease:

struct predref
{
  TestPredicate & p;
  bool operator()(int n) { return p(n); }
  predref(TestPredicate & r) : p(r) { }
};

container.remove_if(predref(pred));