Reputation: 945
Is there a way I can compare to the type of a code object constructed by compile
or __code__
to the actual code object type?
This works fine:
>>> code_obj = compile("print('foo')", '<string>', 'exec')
>>> code_obj
<code object <module> at 0x7fb038c1ab70, file "<string>", line 1>
>>> print(type(code_obj))
code
>>> def foo(): return None
>>> type(foo.__code__) == type(code_obj)
True
But I can't do this:
>>> type(foo.__code__) == code
NameError: name 'code' is not defined
but where do I import code
from?
It doesn't seem to be from code.py. It's defined in the CPython C
file but I couldn't find the Python
interface type for it.
Upvotes: 2
Views: 418
Reputation: 160567
You're after CodeType
which can be found in types
.
>>> from types import CodeType
>>> def foo(): pass
...
>>> type(foo.__code__) == CodeType
True
Note that there's nothing special about it, it just uses type
on a functions __code__
.
Since it's in the standard lib, you can be sure it will work even if some change happens in the way code objects are exposed.
Upvotes: 4