Does std::unordered_map operator[] do zero-initialization for non-exisiting key?

On the site you linked it says:

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

So the int is value-initialized:

The effects of value initialization are:


4) otherwise, the object is zero-initialized

This is why the result is 0.

Depending on which overload we're talking about, std::unordered_map::operator[] is equivalent to []

T& operator[](const key_type& k)
    return try_­emplace(k).first->second;

(the overload taking an rvalue-reference just moves k into try_emplace and is otherwise identical)

If an element exists under key k in the map, then try_emplace returns an iterator to that element and false. Otherwise, try_emplace inserts a new element under the key k, and returns an iterator to that and true []:

template <class... Args>
pair<iterator, bool> try_emplace(const key_type& k, Args&&... args);

Interesting for us is the case of there being no element yet []/6:

Otherwise inserts an object of type value_­type constructed with piecewise_­construct, forward_­as_­tuple(k), forward_­as_­tuple(std​::​forward<Args>(args)...)

(the overload taking an rvalue-reference just moves k into forward_­as_­tuple and, again, is otherwise identical)

Since value_type is a pair<const Key, T> []/2, this tells us that the new map element will be constructed as:

pair<const Key, T>(piecewise_­construct, forward_­as_­tuple(k), forward_­as_­tuple(std​::​forward<Args>(args)...));

Since args is empty when coming from operator[], this boils down to our new value being constructed as a member of the pair from no arguments [pairs.pair]/14 which is direct initialization [class.base.init]/7 of a value of type T using () as initializer which boils down to value initialization [dcl.init]/17.4. Value initialization of an int is zero initialization [dcl.init]/8. And zero initialization of an int naturally initializes that int to 0 [dcl.init]/6.

So yes, your code is guaranteed to return 0…