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Home/ Questions/Q 91873
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Asked: May 10, 20262026-05-10T23:05:53+00:00 2026-05-10T23:05:53+00:00

When I try this code: a, b, c = (1, 2, 3) def test():

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When I try this code:

a, b, c = (1, 2, 3)  def test():     print(a)     print(b)     print(c)     c += 1 test() 

I get an error from the print(c) line that says:

UnboundLocalError: local variable 'c' referenced before assignment 

in newer versions of Python, or

UnboundLocalError: 'c' not assigned 

in some older versions.

If I comment out c += 1, both prints are successful.

I don’t understand: why does printing a and b work, if c does not? How did c += 1 cause print(c) to fail, even when it comes later in the code?

It seems like the assignment c += 1 creates a local variable c, which takes precedence over the global c. But how can a variable "steal" scope before it exists? Why is c apparently local here?


See also Using global variables in a function for questions that are simply about how to reassign a global variable from within a function, and Is it possible to modify a variable in python that is in an outer (enclosing), but not global, scope? for reassigning from an enclosing function (closure).

See Why isn't the 'global' keyword needed to access a global variable? for cases where OP expected an error but didn’t get one, from simply accessing a global without the global keyword.

See How can a name be "unbound" in Python? What code can cause an `UnboundLocalError`? for cases where OP expected the variable to be local, but has a logical error that prevents assignment in every case.

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1 Answer

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  1. 2026-05-10T23:05:54+00:00Added an answer on May 10, 2026 at 11:05 pm

    Python treats variables in functions differently depending on whether you assign values to them from inside or outside the function. If a variable is assigned within a function, it is treated by default as a local variable. Therefore, when you uncomment the line, you are trying to reference the local variable c before any value has been assigned to it.

    If you want the variable c to refer to the global c = 3 assigned before the function, put

    global c 

    as the first line of the function.

    As for python 3, there is now

    nonlocal c 

    that you can use to refer to the nearest enclosing function scope that has a c variable.

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  • added an answer I have no idea why your code doesn't work, but… May 11, 2026 at 10:34 am
  • added an answer If nothing else double-check that you have a newline at… May 11, 2026 at 10:34 am
  • added an answer OK, perhaps I didn't phrase the question very well, or… May 11, 2026 at 10:34 am

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