Will the pair be complete after map insertion?

Question

I am inserting elements in a map avoiding duplicates, so I've written this code

std::unordered_map<size_t, size_t> idToPos;
for (it : list) {
    auto itPos = idToPos.find(it.id());
    if (itPos==idToPos.end()) {
        idToPos[itPos->first] = idToPos.size();
        do_something_else();
    }
}

The question is, if itPos==idToPos.end() is true, will itPos->second yield the same result as idToPos[itPos->first] given idToPos is an unordered_map?

Why do you not just use `idToPos[it.id()] = idToPos.size()`? — Lasersköld, Apr 16 '21 at 09:14
And do you really need to do the check? unordered_map will take care of not getting duplicate ids... — Lasersköld, Apr 16 '21 at 09:16
@Lasersköld: `idToPos.emplace(it.id, idToPos.size());` as existing objects are not overwritten. — Jarod42, Apr 16 '21 at 09:16
@Lasersköld the problem was that getting the id can be costly. However, I see my blunder, since if ```itPos==idToPos.end()``` then looking for ```first``` or ```second``` makes no sense. — Maxence1402, Apr 16 '21 at 09:22
@Jarod42 Shouldn't it be ```try_emplace``` rather than ```emplace```? — Maxence1402, Apr 16 '21 at 09:38
@Maxence1402, for `std::size_t` key and value it doesn't really matter. — Evg, Apr 16 '21 at 10:37

YSC · Answer 1 · 2021-04-16T09:24:26.993

1

When itPos==idToPos.end(), itPos->first is undefined, which leads to Undefined Behaviour.

You want

std::unordered_map<size_t, size_t> idToPos;
for (it : list) {
    auto itPos = idToPos.find(it.id());
    if (itPos==idToPos.end()) idToPos[it.id()] = idToPos.size();
}

edited Apr 16 '21 at 09:24

answered Apr 16 '21 at 09:22

YSC

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What does UB mean? – Maxence1402 Apr 16 '21 at 09:23
My bad, UB means Undefined Behaviour. Let me update the answer. – YSC Apr 16 '21 at 09:25
Is there any way to avoid getting the id twice? Depending on the element it can be costly. I guess I can still use a new variable – Maxence1402 Apr 16 '21 at 09:28
`.emplace()` will handle duplicates, there is no need to use `.find` (searching the map twice, first time in `find`, second time in `[]`). – Evg Apr 16 '21 at 09:33
@Evg I did not specify though that I want to do something else if and only if the key isn't present. I read that ```emplace``` will return a pair whose second element is a boolean telling if the key is absent, is that it? – Maxence1402 Apr 16 '21 at 09:45
@Maxence1402, that's true. – Evg Apr 16 '21 at 10:36

Will the pair be complete after map insertion?

1 Answers1