Each of next() and list() iterates over generator with mutable object differently

Question

def generator(dct):
    for i in range(3):
        dct[\'a\'] = i
        yield dct

g = generator({\'a\': None})
next(g) # -> {\'a\': 0}
next(g) # -> {\'a\': 1}
n         


        

Answer 1

You are looking at intermediate print results. You are modifying a single dictionary, and sharing references to it. You can see the intermediary steps, but that doesn't mean the result is different.

Store a reference to the object returned from next() and they'll all be the same:

g = generator({'a': None})
first = next(g)
second = next(g)
third = next(g)
print(first, second, third) # -> {'a': 2}  {'a': 2}  {'a': 2}

first, second and third are all references to the same object:

>>> first is second and second is third
True

You'd see the same thing if you did this in a regular for loop:

>>> results = []
>>> d = {'a': None}
>>> for i in range(3):
...     d['a'] = i
...     print(d)
...     results.append(d)
...
{'a': 0}
{'a': 1}
{'a': 2}
>>> print(results)
[{'a': 2}, {'a': 2}, {'a': 2}]
>>> all(d is elem for elem in results)  # all references in results are the same object
True

The loop prints the dictionary object as it changes each step. The results list contains 3 references to the same object, and each shows the same state once printed at the end.