Std :: unordered_map initialization

Question

When I first get the std :: unordered_map element using the [] operator , it is automatically created. What (if any) are the guarantors of its initialization? (Guaranteed initialization of the value or only for its construction)?

Example:

std::unordered_map<void *, size_t> size; char *test = new char[10]; size[test] += 10; 

Is the size [test] guaranteed to be 10 at the end of this sequence?

Answer 1

Is the size [test] guaranteed to be 10 at the end of this sequence?

Yes. In the last line of your code, the value of size[test] initializes the T() element, or in this case size_t() :

C ++ 11 23.4.4.3 access to the map element [map.access]

T& operator[](const key_type& x) ;

1 Effects: If there is no key equivalent to x on the card, inserts the value__name (x, T () ) into the card.

As for T() , the exact language is somewhat involved, so I will try to bring the corresponding bits:

C ++ 11 8.5.16 The semantics of initializers is as follows.

- If the initializer is (), the object is initialized with a value.

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8.5.7 For value-initialize, an object of type T means:

- if T is a class of a class (possibly cv-qualit) ...

- if T is a (possibly cv-qualified) non-unit class ...

- if T is an array type, then each element is initialized with a value;

- , otherwise the object is initialized to zero.

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8.5.5 For zero initialization, an object or reference of type T means:

- if T is a scalar type (3.9),, the object is set to 0 (zero) , taken as an integral constant expression, converted to T;

Answer 2

What's the difference? Initializing values ​​for objects of type class entails a default construct, so the answer is "both." For the map <K, V> new object will be initialized with V() .

All standard containers initialize new items with a value or direct initialization (the latter, possibly through copy building). It is impossible for the new standard container elements to be in an “uninitialized” state (i.e. there is no mechanism that initializes the elements by default).