std::map::operator[]

Question

I was doing a simple map program but ended up with this question. The c++ doc says this:

Access element If k matches the key of an element in the container, the function returns a reference to its mapped value. If k does not match the key of any element in the container, the function inserts a new element with that key and returns a reference to its mapped value. Notice that this always increases the container size by one, even if no mapped value is assigned to the element (the element is constructed using its default constructor).

The part I don't really get is where it says "the element is constructerd using its default constructor".

I gave it a try and made this

std::map<string, int> m;
m["toast"];

I just wanted to see what value would the mapped element of "toast" be. And it ended up being zero, but, why? does the primitive types have a default constructor? or what is happening?

Answer 1

The statement of "using its default constructor" is confusing. More precisely, for std::map::operator[], if the key does not exist, the inserted value will be value-initialized.

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

For int, it means zero-initialization.

4) otherwise, the object is zero-initialized.

Answer 2

Map values are value-initialized by the operator[], which, for int means zero-initialization.

As defined by the standard (§23.4.4.3):

Effects: If there is no key equivalent to x in the map, inserts value_type(x, T()) into the map.

T() is explained as (§8.5/10):

An object whose initializer is an empty set of parentheses, i.e., (), shall be value-initialized ​

which means (§8.5/8): ​

To value-initialize an object of type T means:

[...]

— otherwise, the object is zero-initialized.

and zero-initialization is defined as (§8.5/6):

To zero-initialize an object or reference of type T means:

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

[...]

_{all quotes taken from n4140}