Question

I am trying to learn how the std::bind works. I wrote the following:

#include <iostream>
#include <algorithm>
#include <functional>


using namespace std::placeholders;

int fun2(int i,int j)
{
   return i+j;
}


int fun(int i)
{
  return i;
}

int main()
{
 std::vector<int> v9={1,2,3,4,5,6,6,7};
 std::transform(v9.begin(),v9.end(),v9.begin(),[](int i){return i;}); //works
 std::transform(v9.begin(),v9.end(),v9.begin(),fun); //works
 std::transform(v9.begin(),v9.end(),v9.begin(),std::bind(fun,_1)); //works
 std::transform(v9.begin(),v9.end(),v9.begin(),std::bind(fun2,_1,_2)); //does not work
}

std::transform also accepts binary operation functions. So I tried to wrote fun2 and use std::bind (last line of the main) but it doesn't work. Can someone give me any example how std::bind use placeholders (2,3 or more)?

Was it helpful?

Solution

The overload of std::transform that takes a binary functor takes four iterators, not three, since it operates on two input ranges, and not one. For example:

#include <iostream>
#include <algorithm>
#include <functional>
#include <iterator>

int fun2(int i,int j)
{
   return i+j;
}

int main()
{
  using namespace std::placeholders;
  std::vector<int> v1={1,2,3,4,5,6,6,7};
  std::vector<int> v2;
  std::transform(v1.begin(), v1.end(), v1.begin(), 
                 std::back_inserter(v2), std::bind(fun2,_1,_2));
  for (const auto& i : v2)
    std::cout << i << " ";
  std::cout << std::endl;
}

Of course, in real life you wouldn't use std::bind here.

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