CODEWITHSUNDEEP
pythonpython-3.x
Sriv
Sriv

Reputation: 405

Unresolved Reference In Python Method

I am trying to find a reason to this Error. The Code Is

def x():
    if b == 1:
        a = a + 4
        print(a)
a = 5
b = 1 

x()

The second time my variable a is used : a = a + 4, an error occur.

I am developing a program which mainly uses this type of assignments.

Upvotes: 3

Views: 15657

Answers (1)

Mike Scotty
Mike Scotty

Reputation: 10782

The assignment a = a + 4 tells the interpreter to use a local version of a instead of the global reference.

Now in the assignment a = a + 4, when the interpreter is looking up the value of the local variable a, the interpreter finds the variable to be undefined and will yield UnboundLocalError: local variable 'a' referenced before assignment.

You can change this behavior by using the global keyword:

def x():
    global a
    if b == 1:
        a = a + 4
        print(a)
        return
a = 5
b = 1
x()

Q: Hold on, why isn't the interpreter complaining about b?

A: Because you're not assigning anything to b within the function, thus the interpreter will use the variable b from the global scope.

However, global variables are not needed for this case. A more elegant version that avoids global variables would be:

def x(a, b):
    if b == 1:
        a = a + 4
    return a
print( x(5, 1) )

Note that this function behaves slightly differently. Your original function did not return anything, while my reworked version of x will always return the value of a, either modified or unmodified, depending on the value of b.

This has the advantage that you can work with the returned value and you can move the print() out of the function.

Upvotes: 14

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