Why is this code giving 'UnboundLocalError: local variable 'num1' referenced before assignment' error?
num1=50
def func():
print(num1)
num1=100
func()
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Because you are redefining `num1` _**inside**_ of `func()`, Python considers `num1` local to `func()`. So when you attempt to print the _global_ `num1`, Python only finds the _local_ `num1` and raises an error because it sees it has not been defined yet. – Christian Dean Jun 20 '17 at 06:15
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But it should take the global value of num1 right? – Saurav.Kumar Jun 20 '17 at 06:48
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No, if you assign to a variable anywhere in a function, the compiler marks it as local, unless you use the `global` directive. – juanpa.arrivillaga Jun 20 '17 at 07:04
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ok thank you got it. – Saurav.Kumar Jun 20 '17 at 07:08
1 Answers
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Another gotcha! of python. This is because of hoisting and variable shadowing. If you have a local and global variable with the same name in a particular scope, the local variable will shadow the global one. Furthermore, declarations are hoisted to the top of their scope.
So your original code will look something like this:
num1=50
def func():
num1 = ... # no value assigned
print(num1)
num1=100
func()
Now, if you try to print num1 without having assigned any value to it, it throws UnboundLocalError since you have not bound any value to the variable at the time you are trying to dereference it.
To fix this, you need to add the global keyword to signify that num1 is a global variable and not local.
num1=50
def func():
global num1
print(num1)
num1=100
func()
cs95
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@Saurav.Kumar Like I said, declarations are moved to the top of their enclosing scope. – cs95 Jun 20 '17 at 07:11