scope of globals, locals, use with exec()

I’m having a hard time understanding the scope of globals(), locals() in Python. Consider the following code:

def f1(glob, loc):
    exec("b = 5", glob, loc)


f1(globals(), locals())
print(b + 1)

This passes globals and locals to f1 where the variable b = 5 is defined, and sure enough it’s present after the function call in the main scope. The code correctly prints

6

To my surprise, when doing the same in the scope of a function, f2 here, it fails:

def f1(glob, loc):
    exec("b = 5", glob, loc)


def f2():
    f1(globals(), locals())
    print(b + 1)


f2()

Traceback (most recent call last):
  File "/tmp/q.py", line 10, in <module>
    f2()
  File "/tmp/q.py", line 7, in f2
    print(b + 1)
NameError: name 'b' is not defined

Why is that? Is it possible to generalize f1 to allow it to be called from f2 as well?

>Solution :

In your first example, locals is the globals dict, because you’re on a module level.

You can check that with print(id(globals()), id(locals()))

In your second example, locals is a different dict in the scope of the function.

So to make example 2 work, either use:

def f1(glob, loc):
    exec("b = 5", glob, loc)
def f2():
    f1(globals(), globals())
    print(b + 1)
f2()

or

def f1(glob, loc):
    exec("b = 5", glob, glob)
def f2(loc):
    f1(globals(), loc)
    print(b + 1)
f2(locals())

So this isn’t really a matter of exec() but of in which scope you’re calling locals().

And ofc it’s also a matter of what the code in exec does, because an assignment changes the scope of the name to the current (local) scope, unless global/nonlocal is involved. That’s why b=5 is executed in the local scope – except that in your first example, the "local" scope is the global scope.