Re: member iterator - request for comment

Paul Bibbings <>
Mon, 21 Jun 2010 17:35:06 CST
Mikosz <> writes:


Recently I've met the need to perform operations like this quite

struct A {
   int val;

std::vector<A> as;
std::set<int> vals;
std::vector<A>::const_iterator it, end = as.end();
for (it = as.begin(); it != end; ++it) {

that is, to perform some action on one of the members of each of the
collection's elements. I couldn't find any solutions within the STL,

    #include <vector>
    #include <set>
    #include <algorithm>

    struct A {
       A(int v) : val(v) { } // added for convenience here
       int val;

    struct val_functor {
       int operator()(A a) { return a.val; }

    int main()
       std::vector<A> as;
       std::set<int> vals;
       for (int i = 0; i < 10; i++) as.push_back(i);

                      std::inserter(vals, vals.begin()),

so I've created my own MemberIterator class. The code is:

template<class Iterator, class PointerToMember, class Member>
class MemberIterator : public Iterator {

     MemberIterator(Iterator it, PointerToMember ptr) : Iterator(it),
ptr_(ptr) {

     Member& operator*() {
         return Iterator::operator*().*ptr_;


     PointerToMember ptr_;


template<class Member, class Iterator, class PointerToMember>
MemberIterator<Iterator, PointerToMember, Member> makeMemberIterator(
         Iterator it, PointerToMember ptr) {
     return MemberIterator<Iterator, PointerToMember, Member>(it, ptr);

Given this solution, I can write:

std::copy(makeMemberIterator<int>(as.begin(), &A::val),
makeMemberIterator<int>(as.end(), &A::val), std::back_inserter(vals));

and it works just fine.

I wouldn't have thought that the above would work at all. Given that
your vals is of type std::set<int>, I would have expected the above line
of code to fail on std::set not defining push_back.

I would like to know your opinion on
1. whether this class is implemented correctly or how I could make it
2. whether I should attempt to write stuff like this or use some STL
provided solution that I'm not aware of

Prefer the second option in 2. which, effectively, makes 1. moot.


Paul Bibbings

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