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Home/ Questions/Q 675599
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T00:50:08+00:00 2026-05-14T00:50:08+00:00

When I try this code: dict_a = dict_b = dict_c = {} dict_c[‘hello’] =

  • 0

When I try this code:

dict_a = dict_b = dict_c = {}
dict_c['hello'] = 'goodbye'

print(dict_a)
print(dict_b)
print(dict_c)

I expected that it would just initialise the dict_a, dict_b and dict_c dictionaries, and then assign a key in dict_c, resulting in

{}
{}
{'hello': 'goodbye'}

But it seems to have a copy-through effect instead:

{'hello': 'goodbye'}
{'hello': 'goodbye'}
{'hello': 'goodbye'}

Why?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T00:50:09+00:00Added an answer on May 14, 2026 at 12:50 am

    This is because in Python, variables (names) are just references to individual objects. When you assign dict_a = dict_b, you are really copying a memory address (or pointer, if you will) from dict_b to dict_a. There is still one instance of that dictionary.

    To get the desired behavior, use either the dict.copy method, or use copy.deepcopy if your dict may have nested dicts or other nested objects.

    >>> a = {1:2}
    >>> b = a.copy()
    >>> b
    {1: 2}
    >>> b[3] = 4
    >>> a
    {1: 2}
    >>> b
    {1: 2, 3: 4}
    >>> 
    
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