Question

What a title, suppose I have a map like this:

std::map<int, int> m;

and if I write the following

cout<<m[4];

• What will be the result (0, uninitialized, compiler specific)?
• Does the same applies for pointers (i.e. std::map)?

EDIT:
To clarify, in this question I am seeking for standard behavior.

Thanks in advance,
Cem

Answer 1

The value will be what the default constructor for the type creates, because new spots are filled using T(). For int, this is 0. You can see this for yourself:

#include <iostream>

using namespace std;

int main() {
    cout << int() << endl; // prints 0
}

Initializing a type like with empty parentheses like this is called value initialization (see ildjarn’s comment below).