When I access an element in std::unordered_map using operator [] for the first time, it is automatically created. What (if any) are guarantees about its initialization? (It is guaranteed to be value initialized, or only to be constructed)? Example:
std::unordered_map size; char *test = new char[10]; size[test] += 10;
Is size[test] guaranteed to be 10 at the end of this sequence?
asked Jan 20, 2012 at 14:53
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Is size[test] guaranteed to be 10 at the end of this sequence?
Yes. In the last line of your code, size[test] value-initializes the element to T() , or in this case size_t() :
C++11 23.4.4.3 map element access [map.access]
T& operator[](const key_type& x) ;
1 Effects: If there is no key equivalent to x in the map, inserts value_type(x, T()) into the map.
As to T() , the exact language is a somewhat involved, so I'll try to quote the relevant bits:
C++11 8.5.16 The semantics of initializers are as follows.
— If the initializer is (), the object is value-initialized.
8.5.7 To value-initialize an object of type T means:
— if T is a (possibly cv-qualified) class type .
— if T is a (possibly cv-qualified) non-union class type .
— if T is an array type, then each element is value-initialized;
— otherwise, the object is zero-initialized.
8.5.5 To zero-initialize an object or reference of type T means:
— if T is a scalar type (3.9), the object is set to the value 0 (zero), taken as an integral constant expression, converted to T;