undefined variable does not raise error if in scope of iteration with that name defined

Question

Using CPython 3.8.2+ (984a5, the following code executes without raising an error. The result is a dictionary which prints as if it has values that are instances with the same value stored.

class Foo(object):
    def __init__(self, name):
        self.name = name

    def __repr__(self):
        return name  # this is undefined (missing `self.`)


optable = dict()
for name in ['a', 'b']:
    optable[name] = Foo(name)
print(optable)
print(optable['a'].name)
print(optable['b'].name)

This script prints

{'a': b, 'b': b}
a
b

Unexpectedly, both representations are printed, and are "b".

Could this be a CPython bug?

`name` is pulled from the global scope, it's defined and assigned by the `for` loop. Try `for item in ['a', 'b']:` instead. — metatoaster, Apr 20 '20 at 04:37
Additional [discussion](https://stackoverflow.com/questions/3611760/scoping-in-python-for-loops). — metatoaster, Apr 20 '20 at 04:39
`name` is *not* undefined, Python scoping rules look first at local, then enclosing, and finally global (and built-in) namespace. `name` exists in the global namespace — juanpa.arrivillaga, Apr 20 '20 at 06:02

score 0 · Accepted Answer · answered Apr 20 '20 at 04:42

As said in the comments, make sure to return the name that belongs to each instance of Foo

    def __repr__(self):
        return self.name

to prevent returning the global name instead.

Also, make sure to use the correct method to represent the object

undefined variable does not raise error if in scope of iteration with that name defined

1 Answers1