Why can't for_each modify its functor argument?

Question

http://www.cplusplus.com/reference/algorithm/for_each/
Unary function taking an element in the range as argument. This can either be a pointer to a function or an object whose class overloads operator(). Its return value, if any, is ignored.

According to this article, I expected that for_each actually modifies the object given as its third argument, but it seems like for_each operates on a temporary object, and doesn't even modify the object given to it.

So, why is it implemented in that way? It seems much less useful. Or did I misunderstand something and my code below contains errors?

#include <iostream>
#include <vector>
#include <algorithm>

template <class T> struct Multiplicator{
    T mresult;
  public:
    const T& result() const{return mresult;}
    Multiplicator(T init_result = 1){
      mresult = init_result;
    }
    void operator()(T element){
      mresult *= element;
      std::cout << element << " "; // debug print
    }
};

int main()
{
    std::vector<double> vec;
    vec.push_back(1);
    vec.push_back(2);
    vec.push_back(3);
    Multiplicator<double> multiply;
    std::for_each(vec.begin(),vec.end(),multiply);
    std::cout << "\nResult: " << multiply.result() << std::endl;
    return 0;
}

Expected output:

1 2 3 Result: 6

But got following output:

1 2 3 Result: 1

How about changing your question title? Might make it easier for others to find. — BeeBand, Jan 20 '10 at 16:04

score 16 · Accepted Answer · answered Jan 20 '10 at 14:37

16

The function object is taken by value. for_each returns the function object, so if you change it to:

multiply = std::for_each(vec.begin(),vec.end(),multiply);

you get the expected output.

answered Jan 20 '10 at 14:37

James McNellis

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Generally for functors, and because they are supposed to be lightweight and copyable, I cheat and use an attribute that references a variable passed into the constructor. Thus any copy is stricly equivalent and refers to the same item. Of course, it makes initialization a bit awkward (declare variable, pass it to functor, read from variable), but it works fine and does not require any particular implementation propriety (like not discarding the just modified functor). – Matthieu M. Jan 20 '10 at 18:41
Lemme give you the first gold STL badge – Johannes Schaub - litb Jul 24 '11 at 17:46
@Johannes: Thank you, but I have to answer another 36 questions tagged [stl] first :-) (the requirement is now at least 1,000 upvotes and at least 200 answers; I remember when they changed it a year ago and I lost my gold [c] badge for a couple months.) – James McNellis Jul 24 '11 at 21:06

score 10 · Answer 2 · answered Jan 20 '10 at 14:48

While James is correct, using std::accumulate with std::multiplies would be more correct, probably:

#include <iostream>
#include <functional>
#include <numeric>
#include <vector>

int main(void)
{
    std::vector<double> vec;
    vec.push_back(1);
    vec.push_back(2);
    vec.push_back(3);

    double result = std::accumulate(vec.begin(), vec.end(),
                                    1.0, std::multiplies<double>());

    std::cout << "\nResult: " << result << std::endl;

}

With your for_each version, you don't really need to copy the functor again, rather:

double result = std::for_each(vec.begin(), vec.end(), multiply).result();

Or C++0x, for fun:

double result = 1;
std::for_each(vec.begin(), vec.end(), [&](double pX){ result *= pX; });

What is this, "suggest the right tool for the job day?" :-) +1 — James McNellis, Jan 20 '10 at 15:32
@James: 8) Unfortunately, the quality of my answers decreases exponentially as the day continues. :p — GManNickG, Jan 20 '10 at 15:44

score 0 · Answer 3 · answered Jan 20 '10 at 14:52

0

The semantics of For_each dont fit into what you are trying to do. accumulate does exactly what you are trying, use that instead.

answered Jan 20 '10 at 14:52

Yogesh Arora

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Why can't for_each modify its functor argument?

3 Answers3

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