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Home/ Questions/Q 887477
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T13:14:13+00:00 2026-05-15T13:14:13+00:00

foo = foo def bar(foo): foo = bar bar(foo) print foo # foo is

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foo = “foo”
def bar(foo):
foo = “bar”

bar(foo)
print foo

# foo is still "foo"...

foo = {'foo':"foo"}
def bar(foo): 
    foo['foo'] = "bar"

bar(foo)
print foo['foo']

# foo['foo'] is now "bar"?

I have a function that has been inadvertently over-writing my function parameters when I pass a dictionary. Is there a clean way to declare my parameters as constant or am I stuck making a copy of the dictionary within the function?

Thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-15T13:14:13+00:00Added an answer on May 15, 2026 at 1:14 pm

    In a case like this, you’d have to copy the dictionary if you want to change it and keep the changes local to the function.

    The reason is that, when you pass a dictionary into your second bar function, Python only passes a reference to the dictionary. So when you modify it inside the function, you’re modifying the same object that exists outside the function. In the first bar function, on the other hand, you assign a different object to the name foo inside the function when you write foo = "bar". When you do that, the name foo inside the function starts to refer to the string "bar" and not to the string "foo". You’ve changed which object the name refers to. But foo inside the function and foo outside the function are different names, so if you change the object labeled by foo inside the function, you don’t affect the name foo outside the function. So they refer to different objects.

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