Does std::map automatically zero values?

Question

If I write:

#include <map>

int main()
{
    std::map<int, double> q;
    q[3] += 4;
    return 0;
}

Can I be assured q[3] is 4, and not that q[3] is 4 + (some random uninitialized garbage from memory)?

downvoted because any serious doc for `map::operator[]` should cover this — 463035818_is_not_an_ai, Feb 01 '18 at 16:32
[the documentation](http://en.cppreference.com/w/cpp/container/map/operator_at) says *If an insertion is performed, the mapped value is value-initialized (default-constructed for class types, zero-initialized otherwise) and a reference to it is returned.* Seems pretty clear to me. — Kevin, Feb 01 '18 at 16:34
Even more than that, std::vector(100) would be a vector of a hundred zeroes. — bipll, Feb 01 '18 at 16:35
@Bathsheba If you look into example for http://en.cppreference.com/w/cpp/container/map/operator_at it explicitly mentions that, I doubt it is buried — Slava, Feb 01 '18 at 16:39
@Slava: Let's resist this becoming some kind of bragging contest. Answering questions well is the best way to show off, should you feel the need to. — Bathsheba, Feb 01 '18 at 16:41

Fred Larson · Accepted Answer · 2018-02-01T17:22:22.150

6

Quoting from cppreference.com:

If an insertion is performed, the mapped value is value-initialized (default-constructed for class types, zero-initialized otherwise) and a reference to it is returned.

So you can count on the value being initialized to zero.

EDIT:

The quote above refers to the situation before C++11. The wording for C++11 and later is harder to understand, but here is what I think is the operative sentence:

When the default allocator is used, this results in the key being copy constructed from key and the mapped value being value-initialized.

The same parenthetical comment in the first quote also applies to the term "value-initialized" here (see the page on value initialization), so it will still zero initialize values of primitive types.

edited Feb 01 '18 at 17:22

answered Feb 01 '18 at 16:34

Fred Larson

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OK, so zero initialized if no default constructor exists. Perfect. – Carbon Feb 01 '18 at 16:34
A formal standard reference would be the icing on the cake here, but +1 anyway – Bathsheba Feb 01 '18 at 16:34
3

@Carbon, not sure thats what it actually says. IF there is no default constructor for an object i expect it to fail – pm100 Feb 01 '18 at 16:36
its not zero initialized, but it is default initialized. If no default constructor exists you cannot use `operator[]`, though you can still use the map – 463035818_is_not_an_ai Feb 01 '18 at 16:38
@FrançoisAndrieux: Do feel cree to edit cppreference - I have on a couple of occasions. – Bathsheba Feb 01 '18 at 16:42
3

@Bathsheba My comment is not meant to indicate an error with cppreference. I meant to say the passage quoted in the answer is taken from a block labeled (until C++11). – François Andrieux Feb 01 '18 at 16:43
@FrançoisAndrieux: That's reassuring; cppreference is *extremely* well-maintained. – Bathsheba Feb 01 '18 at 16:45
@FrançoisAndrieux: Edited. Is that better? – Fred Larson Feb 01 '18 at 17:22

score 4 · Answer 2 · answered Feb 01 '18 at 16:32

4

If q[3] does not exist in the map, then that record is created for you and the value is initialised to zero.

Your code is safe.

answered Feb 01 '18 at 16:32

Bathsheba

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Does std::map automatically zero values?

2 Answers2