179. Comparison of const_iterators to iterators doesn't work

Section: 24.2 [container.requirements] Status: CD1 Submitter: Judy Ward Opened: 1998-07-02 Last modified: 2016-01-28 10:19:27 UTC

Priority: Not Prioritized

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Currently the following will not compile on two well-known standard library implementations:

#include <set>
using namespace std;

void f(const set<int> &s)
  set<int>::iterator i;
  if (i==s.end()); // s.end() returns a const_iterator

The reason this doesn't compile is because operator== was implemented as a member function of the nested classes set:iterator and set::const_iterator, and there is no conversion from const_iterator to iterator. Surprisingly, (s.end() == i) does work, though, because of the conversion from iterator to const_iterator.

I don't see a requirement anywhere in the standard that this must work. Should there be one? If so, I think the requirement would need to be added to the tables in section 24.1.1. I'm not sure about the wording. If this requirement existed in the standard, I would think that implementors would have to make the comparison operators non-member functions.

This issues was also raised on comp.std.c++ by Darin Adler.  The example given was:

bool check_equal(std::deque<int>::iterator i,
std::deque<int>::const_iterator ci)
return i == ci;

Comment from John Potter:

In case nobody has noticed, accepting it will break reverse_iterator.

The fix is to make the comparison operators templated on two types.

    template <class Iterator1, class Iterator2>
    bool operator== (reverse_iterator<Iterator1> const& x,
                     reverse_iterator<Iterator2> const& y);

Obviously: return x.base() == y.base();

Currently, no reverse_iterator to const_reverse_iterator compares are valid.

BTW, I think the issue is in support of bad code. Compares should be between two iterators of the same type. All std::algorithms require the begin and end iterators to be of the same type.

Proposed resolution:

Insert this paragraph after 24.2 [container.requirements] paragraph 7:

In the expressions

    i == j
    i != j
    i < j
    i <= j
    i >= j
    i > j
    i - j

Where i and j denote objects of a container's iterator type, either or both may be replaced by an object of the container's const_iterator type referring to the same element with no change in semantics.

[post-Toronto: Judy supplied a proposed resolution saying that iterator and const_iterator could be freely mixed in iterator comparison and difference operations.]

[Redmond: Dave and Howard supplied a new proposed resolution which explicitly listed expressions; there was concern that the previous proposed resolution was too informal.]


The LWG believes it is clear that the above wording applies only to the nested types X::iterator and X::const_iterator, where X is a container. There is no requirement that X::reverse_iterator and X::const_reverse_iterator can be mixed. If mixing them is considered important, that's a separate issue. (Issue 280.)