Enum variable default value?

Question

The question is simple:

#include <iostream>

enum SomeEnum {  
    EValue1 = 1,  
    EValue2 = 4
};

int main() {
    SomeEnum enummy;
    std::cout << (int)enummy;
}

What will be the output?

Note: This is not an interview, this is code inherited by me from previous developers. Streaming operator here is just for example, actual inherited code doesn't have it.

Answer 1

The program has Undefined Behavior . The value of enummy is indeterminate. Conceptually there is no difference between your code and the following code:

int main() {
   int i;          //indeterminate value
   std::cout << i; //undefined behavior
};

If you had defined your variable at namespace scope, it would be value initialized to 0.

enum SomeEnum {  
    EValue1 = 1,  
    EValue2 = 4,  
};
SomeEnum e; // e is 0
int i;      // i is 0

int main()
{
    cout << e << " " << i; //prints 0 0 
}

Don't be surprised that e can have values different from any of SomeEnum 's enumerator values. Each enumeration type has an underlying integral type(such as int , short , or long ) and the set of possible values of an object of that enumeration type is the set of values that the underlying integral type has. Enum is just a way to conveniently name some of the values and create a new type, but you don't restrict the values of your enumeration by the set of the enumerators' values.

Update: Some quotes backing me:

To zero-initialize an object of type T means:
— if T is a scalar type (3.9), the object is set to the value of 0 (zero) converted to T;

Note that enumerations are scalar types.

To value-initialize an object of type T means:
— if T is a class type blah blah
— if T is a non-union class type blah blah
— if T is an array type, then blah blah — otherwise, the object is zero-initialized

So, we get into otherwise part. And namespace-scope objects are value-initialized