Use of locals() inside a function leads to unexpected behavior. For an example, the following code snippet
def make_tables():
schema = dj.schema('test', locals())
@schema
class A():
definition = """
id: int # id
"""
@schema
class B():
definition = """
-> A
"""Calling the fucntion make_tables() will result in name resolution error for the foreign key reference -> A inside the definition of B despite the fact that the following works just fine:
schema = dj.schema('test', locals())
@schema
class A():
definition = """
id: int # id
"""
@schema
class B():
definition = """
-> A
"""The problem arise due to the fact that, for some reasons, the dictionary returned by the locals() do not stay updated when used inside a function. This can be demonstrated by the following code:
def test_locals():
names = locals()
print(names.keys()) # will print out []
# add new variables
a = 10
b = 5
print(names.keys()) # still print out []!
print(locals().keys()) # ['names', 'a', 'b']
print(names.keys()) # ['names', 'a', 'b']This makes the use of locals() highly unreliable inside a function.
Use of
locals()inside a function leads to unexpected behavior. For an example, the following code snippetCalling the fucntion
make_tables()will result in name resolution error for the foreign key reference-> Ainside the definition ofBdespite the fact that the following works just fine:The problem arise due to the fact that, for some reasons, the dictionary returned by the
locals()do not stay updated when used inside a function. This can be demonstrated by the following code:This makes the use of
locals()highly unreliable inside a function.