Why is Python's 'exec' behaving oddly?

Question

Why does the following code work while the code after it breaks?

I'm not sure how to articulate my question in english, so I attached the smallest code I could come up with to highlight my problem.

(Context: I'm trying to create a terminal environment for python, but for some reason the namespaces seem to be messed up, and the below code seems to be the essence of my problem)

No errors:

d={}
exec('def a():b',d)
exec('b=None',d)
exec('a()',d)

Errors:

d={}
exec('def a():b',d)
d=d.copy()
exec('b=None',d)
d=d.copy()
exec('a()',d)

Answer 1

It is because the d does not use the globals provided by exec; it uses the mapping to which it stored the reference in the first exec. While you set 'b' in the new dictionary, you never set b in the globals of that function.

>>> d={}
>>> exec('def a():b',d)
>>> exec('b=None',d)
>>> d['a'].__globals__ is d
True
>>> 'b' in d['a'].__globals__
True

vs

>>> d={}
>>> exec('def a():b',d)
>>> d = d.copy()
>>> exec('b=None',d)
>>> d['a'].__globals__ is d
False
>>> 'b' in d['a'].__globals__
False

---

If exec didn't work this way, then this too would fail:

mod.py

b = None
def d():
    b

main.py

from mod import d
d()

A function will remember the environment where it was first created.

---

It is not possible to change the dictionary that an existing function points to. You can either modify its globals explicitly, or you can make another function object altogether:

from types import FunctionType

def rebind_globals(func, new_globals):
    f = FunctionType(
        code=func.__code__,
        globals=new_globals,
        name=func.__name__,
        argdefs=func.__defaults__,
        closure=func.__closure__
    )
    f.__kwdefaults__ = func.__kwdefaults__
    return f


def foo(a, b=1, *, c=2):
    print(a, b, c, d)


# add __builtins__ so that `print` is found...    
new_globals = {'d': 3, '__builtins__': __builtins__}
new_foo = rebind_globals(foo, new_globals)
new_foo(a=0)