JuniorTheoryCommon

What does the program print?

Python

def double(x):
    return x * 2


print(double(double(3)) + 1)

Expecting 7, as if the outer double were skipped: the inner call returns 6, the outer call doubles it to 12, and only then + 1 gives 13.
Expecting 14, as if + 1 ran before the outer double: + 1 stands outside both calls, so it is added to the finished result 12.