Local Variable Referenced Before Assignment Python Error Message

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If you're closely following the Python tag on StackOverflow, you'll notice that the same question comes up at least once a week. The question goes on like this:

x = 10deffoo(): x += 1print x foo()

Why, when run, this results in the following error:

Traceback (most recent call last): File "unboundlocalerror.py", line 8, in <module> foo() File "unboundlocalerror.py", line 4, in foo x += 1 UnboundLocalError: local variable 'x' referenced before assignment

There are a few variations on this question, with the same core hiding underneath. Here's one:

lst = [1, 2, 3] deffoo(): lst.append(5) # OK#lst += [5] # ERROR here foo() print lst

Running the statement successfully appends 5 to the list. However, substitute it for , and it raises , although at first sight it should accomplish the same.

Although this exact question is answered in Python's official FAQ (right here), I decided to write this article with the intent of giving a deeper explanation. It will start with a basic FAQ-level answer, which should satisfy one only wanting to know how to "solve the damn problem and move on". Then, I will dive deeper, looking at the formal definition of Python to understand what's going on. Finally, I'll take a look what happens behind the scenes in the implementation of CPython to cause this behavior.

The simple answer

As mentioned above, this problem is covered in the Python FAQ. For completeness, I want to explain it here as well, quoting the FAQ when necessary.

Let's take the first code snippet again:

x = 10deffoo(): x += 1print x foo()

So where does the exception come from? Quoting the FAQ:

This is because when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope.

But is similar to , so it should first read , perform the addition and then assign back to . As mentioned in the quote above, Python considers a variable local to , so we have a problem - a variable is read (referenced) before it's been assigned. Python raises the exception in this case [1].

So what do we do about this? The solution is very simple - Python has the global statement just for this purpose:

x = 10deffoo(): global x x += 1print x foo()

This prints , without any errors. The statement tells Python that inside , refers to the global variable , even if it's assigned in .

Actually, there is another variation on the question, for which the answer is a bit different. Consider this code:

defexternal(): x = 10definternal(): x += 1print(x) internal() external()

This kind of code may come up if you're into closures and other techniques that use Python's lexical scoping rules. The error this generates is the familiar . However, applying the "global fix":

defexternal(): x = 10definternal(): global x x += 1print(x) internal() external()

Doesn't help - another error is generated: . Python is right here - after all, there's no global variable named , there's only an in . It may be not local to , but it's not global. So what can you do in this situation? If you're using Python 3, you have the keyword. Replacing by in the last snippet makes everything work as expected. is a new statement in Python 3, and there is no equivalent in Python 2 [2].

The formal answer

Assignments in Python are used to bind names to values and to modify attributes or items of mutable objects. I could find two places in the Python (2.x) documentation where it's defined how an assignment to a local variable works.

One is section 6.2 "Assignment statements" in the Simple Statements chapter of the language reference:

Assignment of an object to a single target is recursively defined as follows. If the target is an identifier (name):

  • If the name does not occur in a global statement in the current code block: the name is bound to the object in the current local namespace.
  • Otherwise: the name is bound to the object in the current global namespace.

Another is section 4.1 "Naming and binding" of the Execution model chapter:

If a name is bound in a block, it is a local variable of that block.

[...]

When a name is used in a code block, it is resolved using the nearest enclosing scope. [...] If the name refers to a local variable that has not been bound, a UnboundLocalError exception is raised.

This is all clear, but still, another small doubt remains. All these rules apply to assignments of the form which clearly bind to . But the code snippets we're having a problem with here have the assignment. Shouldn't that just modify the bound value, without re-binding it?

Well, no. and its cousins (, , etc.) are what Python calls "augmented assignment statements" [emphasis mine]:

An augmented assignment evaluates the target (which, unlike normal assignment statements, cannot be an unpacking) and the expression list, performs the binary operation specific to the type of assignment on the two operands, and assigns the result to the original target. The target is only evaluated once.

An augmented assignment expression like can be rewritten as to achieve a similar, but not exactly equal effect. In the augmented version, is only evaluated once. Also, when possible, the actual operation is performed in-place, meaning that rather than creating a new object and assigning that to the target, the old object is modified instead.

With the exception of assigning to tuples and multiple targets in a single statement, the assignment done by augmented assignment statements is handled the same way as normal assignments. Similarly, with the exception of the possible in-place behavior, the binary operation performed by augmented assignment is the same as the normal binary operations.

So when earlier I said that is similar to, I wasn't telling all the truth, but it was accurate with respect to binding. Apart for possible optimization, counts exactly as when binding is considered. If you think carefully about it, it's unavoidable, because some types Python works with are immutable. Consider strings, for example:

x = "abc" x += "def"

The first line binds to the value "abc". The second line doesn't modify the value "abc" to be "abcdef". Strings are immutable in Python. Rather, it creates the new value "abcdef" somewhere in memory, and re-binds to it. This can be seen clearly when examining the object ID for before and after the :

>>> x = "abc" >>> id(x) 11173824 >>> x += "def" >>> id(x) 32831648 >>> x 'abcdef'

Note that some types in Python are mutable. For example, lists can actually be modified in-place:

>>> y = [1, 2] >>> id(y) 32413376 >>> y += [2, 3] >>> id(y) 32413376 >>> y [1, 2, 2, 3]

didn't change after , because the object referenced was just modified. Still, Python re-bound to the same object [3].

The "too much information" answer

This section is of interest only to those curious about the implementation internals of Python itself.

One of the stages in the compilation of Python into bytecode is building the symbol table [4]. An important goal of building the symbol table is for Python to be able to mark the scope of variables it encounters - which variables are local to functions, which are global, which are free (lexically bound) and so on.

When the symbol table code sees a variable is assigned in a function, it marks it as local. Note that it doesn't matter if the assignment was done before usage, after usage, or maybe not actually executed due to a condition in code like this:

x = 10deffoo(): if something_false_at_runtime: x = 20print(x)

We can use the module to examine the symbol table information gathered on some Python code during compilation:

importsymtable code = '''x = 10def foo(): x += 1 print(x)''' table = symtable.symtable(code, '<string>', 'exec') foo_namespace = table.lookup('foo').get_namespace() sym_x = foo_namespace.lookup('x') print(sym_x.get_name()) print(sym_x.is_local())

This prints:

So we see that was marked as local in . Marking variables as local turns out to be important for optimization in the bytecode, since the compiler can generate a special instruction for it that's very fast to execute. There's an excellent article here explaining this topic in depth; I'll just focus on the outcome.

The function in handles variable name references. To generate the correct opcode, it queries the symbol table function . For our , this returns a bitfield with in it. Having seen , generates a . We can see this in the disassembly of :

35 0 LOAD_FAST 0 (x) 3 LOAD_CONST 1 (1) 6 INPLACE_ADD 7 STORE_FAST 0 (x) 36 10 LOAD_GLOBAL 0 (print) 13 LOAD_FAST 0 (x) 16 CALL_FUNCTION 1 19 POP_TOP 20 LOAD_CONST 0 (None) 23 RETURN_VALUE

The first block of instructions shows what was compiled to. You will note that already here (before it's actually assigned), is used to retrieve the value of .

This is the instruction that will cause the exception to be raised at runtime, because it is actually executed before any is done for . The gory details are in the bytecode interpreter code in :

TARGET(LOAD_FAST) x = GETLOCAL(oparg); if (x != NULL) { Py_INCREF(x); PUSH(x); FAST_DISPATCH(); } format_exc_check_arg(PyExc_UnboundLocalError, UNBOUNDLOCAL_ERROR_MSG, PyTuple_GetItem(co->co_varnames, oparg)); break;

Ignoring the macro-fu for the moment, what this basically says is that once is seen, the value of is obtained from an indexed array of objects [5]. If no was done before, this value is still , the branch is not taken [6] and the exception is raised.

You may wonder why Python waits until runtime to raise this exception, instead of detecting it in the compiler. The reason is this code:

x = 10deffoo(): if something_true(): x = 1 x += 1print(x)

Suppose is a function that returns , possibly due to some user input. In this case, binds locally, so the reference to it in is no longer unbound. This code will then run without exceptions. Of course if actually turns out to return , the exception will be raised. Python has no way to resolve this at compile time, so the error detection is postponed to runtime.


UnboundLocalError: local variable referenced before assignment

So that sucks

Howdya fix that?

I’ll tell you. First, a simple example of the problem:

#This is valid python 2.5 code #My global variable: USER_COUNT = 0 #functions: def Main(): AddUser() def AddUser(): print 'There are',USER_COUNT,'users so far' # actually run Main() Main()

^ That program WORKS FINE. Function AddUser() just references the USER_COUNT variable, which was declared as a global (outside of any of the function blocks).

Here’s where it goes wrong: when we try to write to or we try to update the value of the global variable

#USER_COUNT is a GLOBAL variable USER_COUNT = 0 def Main(): AddUser() def AddUser(): USER_COUNT = USER_COUNT + 1 print 'There are',USER_COUNT,'users so far' Main()

Then we get: UnboundLocalError: local variable ‘USER_COUNT’ referenced before assignment

So that sucks.

The reason this happens is because AS SOON AS YOU WRITE TO A VARIABLE, that variable is AUTOMATICALLY considered LOCAL to the function block in which its declared. Namely:

#USER_COUNT is a GLOBAL! USER_COUNT = 0 def AddUser(): USER_COUNT = USER_COUNT + 1 print 'There are',USER_COUNT,'users so far'

EVEN THOUGH we declared USER_COUNT as a GLOBAL, the simple act of WRITING TO IT __ANYWHERE__ in the function scuzzles-up the “globalness” of the USER_COUNT variable, and like, automatically makes ANY use of USER_COUNT refer to a LOCAL VARIABLE inside of AddUser().

So howdya fix it?

Easy! You do this:

#global USER_COUNT = 0 def Main(): AddUser() def AddUser(): global USER_COUNT ######!!! IMPORTANT !!! Make sure # to use the GLOBAL version of USER_COUNT, not some # locally defined copy of that. I think this # might be a python feature to stop functions from # clobbering the global variables in a program USER_COUNT = USER_COUNT + 1 print 'There are',USER_COUNT,'users so far' Main()

This is an ok-nice feature that might stop a program’s functions from clobbering the globals (since you really have this “just use it” attitude to a variable and you may have no clue that you’re clobbering a global), but really it might be nice if python were more consistent and required use of this global thing for BOTH read/write. Though I guess it could be kinda convenient behavior .. I don’t know yet, haven’t programmed in python for long enough.

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