UnboundLocalError: local variable referenced before assignment

Question

I have read similar questions on this topic such as from here, here, and here, and others on scope, but I still cannot understand why answers do not explain these test cases.

First see this:

def take_sum(a, b, c):
    return a+b+c

def main():
    print("take_sum" in globals())
    print(take_sum(1,2,3))

main()

Returns:

True
6

So then this must work:

# TEST 1

def take_sum(a, b, c):
    return a+b+c

def main():
    if "take_sum" not in globals():
        take_sum = lambda a, b, c: a+b+c

    print(take_sum)

main()

WRONG!

Returns: UnboundLocalError: local variable 'take_sum' referenced before assignment

Question 1: How is take_sum referenced before assignment if it exists in the global scope and the if-statement in main evalutes to False?

So let's add an else clause to the if-statement:

# TEST 2

def take_sum(a, b, c):
    return a+b+c

def main():
    if "take_sum" not in globals():
        take_sum = lambda a, b, c: a+b+c

    else:
        global take_sum

    print(take_sum)

main()

Returns: SyntaxError: name 'take_sum' is assigned to before global declaration

Question 2: But how is this possible if the error from TEST 1 says take_sum was referenced but NOT assigned?

Subsequently, switching the clauses in the if-statement works:

# TEST 3

def take_sum(a, b, c):
    return a+b+c

def main():
    if "take_sum" in globals():
        global take_sum

    elif "take_sum" not in globals():
        take_sum = lambda a, b, c: a+b+c

    print(take_sum)
    print(take_sum(1,2,3))

main()

Returns:

<function take_sum at 0x7fbf8b5bb160>
6

Question 3: Why does switching clauses (compared to TEST 2) work?

Answer 1

Whether a name used in a function is a global variable or a local one is determined at compile time, not at run time. Your functions that cause exceptions are trying to have it both ways, to either access a global, or provide their own local variable replacement, but Python's scoping rules don't allow that. It needs to be local or global, and can't be in a nebulous either/or state.

In your Test 1, the function raises an exception because the compiler saw that the code could assign to take_sum as a local variable, and so it makes all the references to take_sum in the code be local. You can no longer look up the global variable take_sum in the normal way once that determination has been made.

A global statement is in effect a compiler directive to change the assumption that an assignment makes a variable local. Subsequent assignments will be made globally, not locally. It's not something that executes at runtime, which is why your other two test cases are so confusing to you.

Test 2 fails because you're trying to tell the compiler that take_sum is a global after it has already seen some of your code make a local assignment to that name. In Test 3, the global statement comes first, so it makes the assignment (in the other branch!) assign to a global variable. It doesn't actually matter that the global statement was in a different branch than the assignment, the compiler interprets the global statement at compile time, not at runtime when the conditional logic of the ifs and elifs gets handled.

It might help your understanding of what is going on to disassemble some of the main functions you've written using the dis.dis function in the standard library. You'll see that there are two different sets of bytecodes used for loading and storing of variables, LOAD_GLOBAL/STORE_GLOBAL for global variables (used in all your functions to get names like print and globals), and LOAD_FAST/STORE_FAST which are used for local variables (like a, b and c in take_sum). The compiler behavior I talked about above boils down to which bytecode it chooses for each lookup or assignment.

If I rename the main function in Test 1 to test1, here's what I get when I disassemble it:

dis.dis(test1)
  2           0 LOAD_CONST               1 ('take_sum')
              2 LOAD_GLOBAL              0 (globals)
              4 CALL_FUNCTION            0
              6 CONTAINS_OP              1
              8 POP_JUMP_IF_FALSE       18

  3          10 LOAD_CONST               2 (<code object <lambda> at 0x0000019022A05F50, file "<ipython-input-23-0cc3c65f7038>", line 3>)
             12 LOAD_CONST               3 ('test1.<locals>.<lambda>')
             14 MAKE_FUNCTION            0
             16 STORE_FAST               0 (take_sum)

  5     >>   18 LOAD_GLOBAL              1 (print)
             20 LOAD_FAST                0 (take_sum)
             22 CALL_FUNCTION            1
             24 POP_TOP
             26 LOAD_CONST               0 (None)
             28 RETURN_VALUE

Disassembly of <code object <lambda> at 0x0000019022A05F50, file "<ipython-input-23-0cc3c65f7038>", line 3>:
  3           0 LOAD_FAST                0 (a)
              2 LOAD_FAST                1 (b)
              4 BINARY_ADD
              6 LOAD_FAST                2 (c)
              8 BINARY_ADD
             10 RETURN_VALUE

Notice that the lookup of take_sum on line 5 is on byte 20 in the bytecode, where it uses LOAD_FAST. This is the bytecode that causes the UnboundLocalError, since there has been no local assigned if the global function exists.

Now, lets look at Test 3:

dis.dis(test3)
  2           0 LOAD_CONST               1 ('take_sum')
              2 LOAD_GLOBAL              0 (globals)
              4 CALL_FUNCTION            0
              6 CONTAINS_OP              0
              8 POP_JUMP_IF_FALSE       12

  3          10 JUMP_FORWARD            18 (to 30)

  5     >>   12 LOAD_CONST               1 ('take_sum')
             14 LOAD_GLOBAL              0 (globals)
             16 CALL_FUNCTION            0
             18 CONTAINS_OP              1
             20 POP_JUMP_IF_FALSE       30

  6          22 LOAD_CONST               2 (<code object <lambda> at 0x0000019022A43500, file "<ipython-input-26-887b66de7e64>", line 6>)
             24 LOAD_CONST               3 ('test3.<locals>.<lambda>')
             26 MAKE_FUNCTION            0
             28 STORE_GLOBAL             1 (take_sum)

  8     >>   30 LOAD_GLOBAL              2 (print)
             32 LOAD_GLOBAL              1 (take_sum)
             34 CALL_FUNCTION            1
             36 POP_TOP

  9          38 LOAD_GLOBAL              2 (print)
             40 LOAD_GLOBAL              1 (take_sum)
             42 LOAD_CONST               4 (1)
             44 LOAD_CONST               5 (2)
             46 LOAD_CONST               6 (3)
             48 CALL_FUNCTION            3
             50 CALL_FUNCTION            1
             52 POP_TOP
             54 LOAD_CONST               0 (None)
             56 RETURN_VALUE

Disassembly of <code object <lambda> at 0x0000019022A43500, file "<ipython-input-26-887b66de7e64>", line 6>:
  6           0 LOAD_FAST                0 (a)
              2 LOAD_FAST                1 (b)
              4 BINARY_ADD
              6 LOAD_FAST                2 (c)
              8 BINARY_ADD
             10 RETURN_VALUE

This time the lookup of take_sum happens on bytecode 40, and it's a LOAD_GLOBAL (which succeeds since there is a global variable of that name).

Answer 2

Take this example

You're pre-defining a variable reference to be global

def main():
  global take_sum
  if "take_sum" not in globals():
    take_sum = lambda a, b, c: a+b+c
  print(take_sum)

Output: <function main.<locals>.<lambda> at 0x7f08c2c54820>

Remove, the not, 'take_sum' is not defined, as expected

---

Now, create a global one

def take_sum(*args):
    return sum(args)

def main():
  global take_sum
  if "take_sum" not in globals():
    take_sum = lambda a, b, c: a+b+c
  print(take_sum)

Output this time is the outer one.

Remove the not, then the local one

---

So, if you try to make global take_sum in the else statement, then as the error says, assigned to before global declaration, and seems like the global gets evaluated without respect for the conditional