Is value in std::unordered_map<std::string, int> initialised by default or not?

I have a following code:

#include <iostream>
#include <unordered_map>
#include <string>

int main()
{
  std::unordered_map<std::string, int> map;
  std::cout << map["foo"]; // -> 0

  return 0;
}

I am using map["foo"] without initialisation. Is this code valid or UB?

Edit: I tried to compile in godbolt and got 0 output with clang and gcc (trunk version) https://godbolt.org/z/KeWq7nbob.

Solution

When operator[] does not find the value with given key one is inserted into the map. From cppreference:

Inserts value_type(key, T()) if the key does not exist. [...]

T() is Zero-initialization. For int this bullet applies:

If T is a scalar type, the object is initialized to the value obtained by explicitly converting the integer literal 0 (zero) to T.

All is fine in your code. map["foo"] returns a reference to 0 after inserting that 0 together with the key constructed from the string literal "foo" into the map. This is the intended use of std::map::operator[]. If you want to check if an element exists in the map use std::map::find. If you want to find an existing one or insert if none is present, use std::map::insert.