Is the order of iterating through std::map known (and guaranteed by the standard)?

Question

What I mean is - we know that the std::map's elements are sorted according to the keys. So, let's say the keys are integers. If I iterate from std::map::begin() to std::map::end() using a for, does the standard guarantee that I'll iterate consequently through the elements with keys, sorted in ascending order?

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Example:

std::map<int, int> map_;
map_[1] = 2;
map_[2] = 3;
map_[3] = 4;
for( std::map<int, int>::iterator iter = map_.begin();
     iter != map_.end();
     ++iter )
{
    std::cout << iter->second;
}

Is this guaranteed to print 234 or is it implementation defined?

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Real life reason: I have a std::map with int keys. In very rare situations, I'd like to iterate through all elements, with key, greater than a concrete int value. Yep, it sounds like std::vector would be the better choice, but notice my "very rare situations".

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EDIT: I know, that the elements of std::map are sorted.. no need to point it out (for most of the answers here). I even wrote it in my question.
I was asking about the iterators and the order when I'm iterating through a container. Thanks @Kerrek SB for the answer.

Answer 1

Yes, that's guaranteed. Moreover, *begin() gives you the smallest and *rbegin() the largest element, as determined by the comparison operator, and two key values a and b for which the expression !compare(a,b) && !compare(b,a) is true are considered equal. The default comparison function is std::less<K>.

The ordering is not a lucky bonus feature, but rather, it is a fundamental aspect of the data structure, as the ordering is used to determine when two keys are the same (by the above rule) and to perform efficient lookup (essentially a binary search, which has logarithmic complexity in the number of elements).