Question

Hi, there.

I have two files:

a.py:

print('in a')
import b

print('var')
VAR = 1

def p():
    print('{}, {}'.format(VAR, id(VAR)))

if __name__ == '__main__':
    VAR = -1
    p()
    b.p() # Where does this VAR come from?

b.py:

print('in b')
import a

def p():
    a.p()

I don’t understand why there’re two different VARs, which is supposed to the same.

If I move the ‘main’ block to another file, everything works well, i.e, there is only one VAR.

c.py:

import a
import b

if __name__ == '__main__':
    a.VAR = -1
    a.p()
    b.p()

So my question is:

Why do the last two lines of a.py print different results?

Don’t they print the same VAR variable in a.py?

BTW, I’m using python 2.7 on win7.

Thanks.

Answer 1

You might want to read up on global variables? To quote:

If a variable is assigned a new value anywhere within the function’s body, it’s assumed to be a local. If a variable is ever assigned a new value inside the function, the variable is implicitly local, and you need to explicitly declare it as ‘global’.

Edit: to elaborate, here’s what happens (leaving out c.py for clarity):

1. File a.py is executed.
2. File b.py is imported, and in turn imports a.py again.
3. Via b’s import, VAR is defined in with a value of 1. This ends b`s import.
4. The __main__ part in a.py is executed in its own scope; VAR is set to -1 there, and p() is ran: this displays -1, as VAR was just set to.
5. Then b.p() is executed, which in turn runs a.p(). Because VAR from the perspective of b.py (different scope) still has a value of 1, the print statement just outputs 1.

You’ll also notice that the id’s are different in both print statements: this is because the VAR‘s live in a different scope, and are not connected in any way. They are disconnected because __main__ lives in a different, anonymous scope: the scope in which the Python interpreter executes. This is briefly discussed in the docs.