Python exec with a function chain producing NameError

Question

Consider the following script, which uses exec to define two functions, one of which calls the other:

def run_code():
  code = """
def foo():
  print('foo')
  return 1

def bar():
  print('bar calls foo')
  return 1 + foo()

result = bar()
"""

  exec(code, globals(), locals())
  print('Result: {}'.format(locals()['result']))

run_code()

I would expect to see the following output:

bar calls foo
foo
Result: 2

but instead, I get the following output+stacktrace:

bar calls foo
Traceback (most recent call last):
  File "minimal.py", line 17, in <module>
    run_code()
  File "minimal.py", line 14, in run_code
    exec(code, globals(), locals())
  File "<string>", line 10, in <module>
  File "<string>", line 8, in bar
NameError: name 'foo' is not defined

Interestingly, if the content of run_code is moved into the module level, then it works fine. However, if I then replace globals() or locals() with a new empty dictionary, it breaks once again. I also know that putting def foo inside bar's body will make it work.

Why is this error occurring, and what is the proper fix?

(I know that exec is generally frowned upon. I am using it for good reason.)

Modifying `locals()` is undefined behavior, and you're doing that here. — user2357112, Oct 29 '20 at 04:11
@user2357112supportsMonica If I replace `locals()` and `globals()` with new empty dictionaries, the issue persists. — k_ssb, Oct 29 '20 at 04:18

score 3 · Accepted Answer · answered Oct 29 '20 at 04:24

From the documentation:

If provided, locals can be any mapping object. Remember that at module level, globals and locals are the same dictionary. If exec gets two separate objects as globals and locals, the code will be executed as if it were embedded in a class definition.

And class definitions do not create enclosing scope, note, this is why you cannot call a method from another method without using self. So just pass the globals() dictionary. Or pass two of the same dict's to both arguments.

In [4]: def run_code():
   ...:     code = """
   ...: def foo():
   ...:   print('foo')
   ...:   return 1
   ...:
   ...: def bar():
   ...:   print('bar calls foo')
   ...:   return 1 + foo()
   ...:
   ...: result = bar()
   ...: """
   ...:     namespace = {}
   ...:     exec(code, namespace)
   ...:     print('Result: {}'.format(namespace['result']))
   ...:

In [5]: run_code()
bar calls foo
foo
Result: 2

score 0 · Answer 2 · answered Oct 29 '20 at 04:41

0

code = """  
def foo():
  print('foo')
  return 1

def bar():
  global foo;
  print('bar calls foo')
  return 1 + foo()

result = bar()
"""
def run_code():
    exec(code, globals(), locals())
    print('Result: {}'.format(locals()['result']))


run_code()

Output:

bar calls foo
foo
Result: 2

answered Oct 29 '20 at 04:41

Aaj Kaal

1,205
1
9
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1

wait, what the heck - `global` is not supposed to work like that. – user2357112 Oct 29 '20 at 04:55
1

I think this is actually a bug. `global foo` inside `bar` is causing `def foo()` to store `foo` in globals instead of locals, but that `global` declaration is only supposed to affect assignments to `foo` inside `bar`. – user2357112 Oct 29 '20 at 04:58

Python exec with a function chain producing NameError

2 Answers2