How to combine std::copy_if and std::transform?

Question

consider this code snippet : iteration over one container of a first type T1 for creating a second container of a second type T2 applying a transformation function T1->T2 but only for T1 elements verifying a predicate (T1 -> bool )

(is Odd in the following example).

std::vector<int> myIntVector;
myIntVector.push_back(10);
myIntVector.push_back(15);
myIntVector.push_back(30);
myIntVector.push_back(13);

std::vector<std::string> myStringVectorOfOdd;

std::for_each(myIntVector.begin(), myIntVector.end(),
    [&myStringVectorOfOdd](int val)
{
    if (val % 2 != 0)
        myStringVectorOfOdd.push_back(std::to_string(val));

});

What I don't like in this code is the capture on the lambda. Is there a way to combine std::copy_if and std::transform to achieve the same result in a more elegant and concise way ?

Answer 1

Here is a transform_if template that takes the usual input iterator pair, an output iterator and a predicate as well as a transformation function object.

template <class InputIt, class OutputIt, class Pred, class Fct>
void transform_if(InputIt first, InputIt last, OutputIt dest, Pred pred, Fct transform)
{
   while (first != last) {
      if (pred(*first))
         *dest++ = transform(*first);

      ++first;
   }
}

You can use it for your example like the following.

transform_if(myIntVector.cbegin(), myIntVector.cend(),
    std::back_inserter(myStringVectorOfOdd),
    [](int n){ return n % 2 != 0; },
    [](int n){ return std::to_string(n); });

It's not super concise, but filtering and transformation are well separated into to capture-free lambdas, and the algorithm itself idiomatically works on iterators.

As range libraries offer better support for composing algorithms, here is the same based on Boost range:

#include <boost/range/algorithm.hpp>
#include <boost/range/adaptors.hpp>

using boost::adaptors::filtered;

boost::transform(myIntVector | filtered([](int n){ return n % 2 != 0; }),
    std::back_inserter(myStringVectorOfOdd), [](int n){ return std::to_string(n); });