UnboundLocalError on local variable when reassigned after first use


The following code works as expected in both Python 2.5 and 3.0:

a, b, c = (1, 2, 3)

print(a, b, c)

def test():
    print(c)    # (A)
    #c+=1       # (B)

However, when I uncomment line (B), I get an UnboundLocalError: "c" not assigned at line (A). The values of a and b are printed correctly. This has me completely baffled for two reasons:

  1. Why is there a runtime error thrown at line (A) because of a later statement on line (B)?

  2. Why are variables a and b printed as expected, while c raises an error?

The only explanation I can come up with is that a local variable c is created by the assignment c+=1, which takes precedent over the "global" variable c even before the local variable is created. Of course, it doesn"t make sense for a variable to "steal" scope before it exists.

Could someone please explain this behavior?

Answer rating: 244

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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