UnboundLocalError: local variable 'a' referenced before assignment

Question

if execute the following code will show error message:

UnboundLocalError: local variable 'a' referenced before assignment

a = 220.0
b = 4300.0
c = 230.0/4300.0

def fun():
    while (c > a/b):
        a = a + 1
        print a/b

if __name__ == '__main__':
    fun()

but modify to :

a = 220.0
b = 4300.0
c = 230.0/4300.0

def fun():
    aa = a
    bb = b
    while (c > aa/bb):
        aa = aa + 1
        print aa/bb

if __name__ == '__main__':
    fun()

it will fine. Any advice or pointers would be awesome. Thanks a lot!

Now what did you want? Modify the global `a` or create a local `a` shadowing the global one? — Jan Hudec, May 02 '13 at 07:16

Ashwini Chaudhary · Accepted Answer · 2013-05-02T07:19:37.403

10

You can't modify a global variable without using the global statement:

def fun():
    global a 
    while (c > a/b):
        a = a + 1
        print a/b

As soon as python sees an assignment statement like a = a + 1 it thinks that the variable a is local variable and when the function is called the expression c > a/b is going to raise error because a is not defined yet.

edited May 02 '13 at 07:19

answered May 02 '13 at 07:13

Ashwini Chaudhary

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UnboundLocalError: local variable 'a' referenced before assignment

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