std :: map :: erase和迭代器

Question

I have some thing like this code:

map<int, string> m;

m[1] = "a";
m[2] = "b";
m[3] = "a";
m[4] = "a";
m[5] = "e";
m[6] = "f";
m[7] = "g";
m[8] = "h";
m[9] = "i";

for (it1 = src.begin(); it1 != src.end(); ++it1) {

    for (it2 = it1; it2 != src.end(); ++it2) {
        if (it2 == it1) {
            continue;
        }

        if (it2->second == it1->second) {
            fprintf(stderr, "%u\n", it2->first);
            src.erase(it2);
        }
    }
}

I use map , because the elements is not always in this order (1, 2 ...)
So here is the question

In some cases of map values, this code print this

2
3
4
6
7
8
9
5

How it is possible (skipping 5 ), if map sort by container in order 1, 2 ... and so on ?

Answer 1

Your erase loop is off. The typical idiom is this:

for(std::map<K,V>::const_iterator it = v.begin(); it != v.end() /* not hoisted! */; /* no increment */ )
{
  // do something
  if (suitable_condition)
  {
    v.erase(it++);
  }
  else
  {
    ++it;
  }
}

Your code erroneously performs an increment on an invalid iterator (namely it2 after the erase ), which is undefined behaviour.

_{(For certain other container types, erase returns an iterator to the next valid element, so in those cases you would say it = v.erase(it) ; but the details of the best erasing pattern depend on the concrete container type.)}