Python: Variable scope in if-statement

In Java, this is wrong and you cannot even compile:

public class HelloWorld{
     public static void main(String []args) {
        if (2 > 1) {
            int a = 2;
        } else {
            int a = 1;
        }

        System.out.println(a);
     }
}

// Error: cannot find symbol ‘a’

And you are used to such a practice:

public class HelloWorld{
     public static void main(String []args) {
        int a = 0;

        if (2 > 1) {
            a = 2;
        } else {
            a = 1;
        }

        System.out.println(a);
     }
}

// OK
// Output: 2

In python, things are a little different:

if __name__ == '__main__':
    # a = 0  # NOT NECESSARY!
    if 2 > 1:
        a = 2
    else:
        a = 1

    print(a)

# OK
# Output: 2

This is because there are 4 types of scopes in python:

• the innermost scope, which is searched first, contains the local names
• the scopes of any enclosing functions, which are searched starting with the nearest enclosing scope, contains non-local, but also non-global names
• the next-to-last scope contains the current module’s global names
• the outermost scope (searched last) is the namespace containing built-in names

and control blocks like if and while don’t count a innermost scope.

It’s even legal to write:

if __name__ == '__main__':
    if 1 > 2:
        a = 1

    print(a)

# UnboundLocalError: local variable 'a' referenced before assignment

although you’ll get a UnboundLocalError.

The error message suggests that if 1 > 2: a = 1 declares a variable a but does not assign a value to it.