Python AST exec“...未定义”递归错误 function

Question

I came across this error

def test_rec():
    import ast
    exec(compile(ast.fix_missing_locations(ast.parse("""
def fact(n):
    return 1 if n == 0 else n * fact(n - 1)

print(fact(5))
"""), "<string>", "exec")))

This yield this error, which is weird

Traceback (most recent call last):
  File "/Users/gecko/.pyenv/versions/3.9.0/envs/lampy/lib/python3.9/site-packages/nose/case.py", line 198, in runTest
    self.test(*self.arg)
  File "/Users/gecko/code/lampycode/tests/test_let_lang.py", line 6, in test_rec
    exec(compile(ast.fix_missing_locations(ast.parse("""
  File "<string>", line 4, in <module>
  File "<string>", line 3, in fact
NameError: name 'fact' is not defined

If I copy and paste the same code in REPL it works fine

>>> def fact(n):
...     return 1 if n == 0 else n * fact(n - 1)
... 
>>> print(fact(5))
120
>>>

Any ideas?

I could reduce the problem further here is the minimal exempla, this would overflow the stack but it gives me the same not defined error

def test_rec3():
    exec("""
def f():
    f()

f()
""")

--

Second edit, going even further, this only happens inside functions

This works

exec("""
def f(n):
    print("end") if n == 1 else f(n-1)

f(10)""")

But this gives me the same error as above

def foo():
    exec("""
def f(n):
    print("end") if n == 1 else f(n-1)

f(10)""")


foo()

Answer 1

If you use exec with the default locals, then binding local variables is undefined behavior. That includes def , which binds the new function to a local variable.

Also, functions defined inside exec can't access closure variables, which fact would be.

The best way to avoid these problems is to not use exec . The second best way is to provide an explicit namespace:

namespace = {}
exec(whatever, namespace)