UnboundLocalError in python confusing

Question

Could anyone explain the exception the below code. It only works when I change the var sub in the display() to another name. There is no global variable sub as well. So what happened ?

def sub(a, b):
    return a - b

def display():
    sub = sub(2,1) // if change to sub1 or sth different to sub, it works
    print sub

Initially, `sub` is a function. Then, it becomes the return value of a function. So when you say `print sub`, python doesn't know which `sub` you are referring to — inspectorG4dget, Oct 01 '12 at 12:19
@inspectorG4dget: your explanation very straightforward. thanks — Hoan Dang, Oct 01 '12 at 12:23

l4mpi · Answer 1 · 2012-10-01T13:14:32.917

Any variable you assign to inside a scope is treated as a local variable (unless you declare it global, or, in python3, nonlocal), which means it is not looked up in the surrounding scopes.

A simplified example with the same error:

def a(): pass

def b(): a = a()

Now, consider the different scopes involved here:

The global namespace contains a and b.

The function a contains no local variables.

The function b contains an assignment to a - this means it is interpreted as a local variable and shadows the function a from the outer scope (in this case, the global scope). As a has not been defined inside of b before the call, it is an unbound local variable, hence the UnboundLocalError. This is exactly the same as if you had written this:

def b(): x = x()

The solution to this is simple: choose a different name for the result of the sub call.

It is important to note that the order of use and assignment makes no difference - the error would have still happened if you wrote the function like this:

def display():
    value = sub(2,1)         #UnboundLocalError here...
    print value
    sub = "someOtherValue"   #because you assign a variable named `sub` here

This is because the list of local variables is generated when the python interpreter creates the function object.

so the unbound error comes from the call to the function because it is first shadowed by the previous assignement? is it the same if x (local variable) is assigned to any value, and then x (function) is called ? — njzk2, Oct 01 '12 at 12:57
The error is that the your function contains an assignment to `sub` so sub is treated as a local variable. And you try to use this local variable during the initial assignment while it doesn't have a value yet. — l4mpi, Oct 01 '12 at 13:09

score 1 · Answer 2 · edited May 23 '17 at 11:56

This was originally a comment. The OP found this useful as an answer. Therefore, I am re-posting it as an answer

Initially, sub is a function. Then, it becomes the return value of a function. So when you say print sub, python doesn't know which sub you are referring to.

Edit:

First you define a function sub. Now, python knows what sub is.

When you create a variable and try to assign to it (say x = 2), python evaluates the stuff on the right hand side of the = and assigns the value of the evaluation as the value of the stuff on the left hand side of the =. Thus, everything on the right hand side should actually compute.

So if your statement was x = x+1, then x better have a value assigned to it before that line; and the previously defined x has to be of some type compatible with the addition of 1.

But suppose x is a function, and you make a variable called x in some other function, and try to assign to it, a value computed with function x, then this really starts to confuse python about which x you are referring to. This is really an oversimplification of this answer, which does a much better job of explaining variable scope and shadowing in python functions

This is overly simplified - if you said `sub = 1` inside `display` you still shadow the function name, which is perfectly legal. The problem is that the name is *used* and *assigned to* in the same scope. See [this answer](http://stackoverflow.com/a/12092094/1110381) for a more detailed explanation of closures in python. — l4mpi, Oct 01 '12 at 12:30
@l4mpi: fair enough. I really should stop answering questions on Monday mornings before my coffee kicks in. Thank you for the clarification — inspectorG4dget, Oct 01 '12 at 12:57

score 1 · Answer 3 · answered Oct 01 '12 at 13:23

For every variable used, Python determines whether it is a local or a nonlocal variable. Referencing a unknown variable marks it as nonlocal. Reusing the same name as a local variable later is considered a programmers mistake.

Consider this example:

def err():
    print x # this line references x
    x = 3   # this line creates a local variable x
err()

This gives you

Traceback (most recent call last):
  File "asd.py", line 5, in <module>
    err()
  File "asd.py", line 2, in err
    print x # this line references x
UnboundLocalError: local variable 'x' referenced before assignment

What happens is basically that Python keeps track of all references to names in code. When it reads the line print x Python knows that x is a variable from a outer scope (upvalue or global). However, in x = 3 the x is used as a local variable. As this is a inconsistency in the code, Python raises an UnboundLocalError to get the Programmers attention.

score -1 · Answer 4 · answered Oct 01 '12 at 12:41

Python start executing your code and get the function first

def sub(a, b):
    return a - b

So after executing this interpreter get the sub as a function. Now when come to next line it found

def display():
    sub = sub(2,1) // if change to sub1 or sth different to sub, it works
    print sub

so first line sub = sub (2, 1) will convert the sub function to sub variable. From this function you are returning the sub variable. So its create problem.

UnboundLocalError in python confusing

4 Answers4