reporting "cannot access local variable ... where it is not associated with a value" in the wrong place when using +=

[inhahe]

Bug report

Bug description:

>>> i = 1
>>> def a():
...   print(i)
...
>>> a()
1
>>> i = 1
>>> def a():
...   i = 2
...   print(i)
...
>>> a()
2
>>> i = 1
>>> def a():
...   i += 1
...   print(i)
...
>>> a()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in a
UnboundLocalError: cannot access local variable 'i' where it is not associated with a value

Sorry if this isn't actually a bug, but I can't believe this was intentional...it makes no sense.

CPython versions tested on:

3.11, 3.13

Operating systems tested on:

Windows

Edit:
Okay, so I did some more testing and realized that

i = 1
def a():
  i = i + 1
  print(i)

also results in this error, but at the i = i + 1 line and not the print(i) line.
So I guess the problem is just a clash between the local name and the global name.
But it's confusing that when you use += it results in the variable no longer being global OR local and gives an error the next time you try to use it..
In my actual use case, it highlighted the variable in a statement unrelated to the assignment like 10 lines down from the assignment.

[skirpichev]

Why do you think it's a bug?

Documentation says: "If a name is bound in a block, it is a local variable of that block, unless declared as nonlocal or global."

In your last example you just do i = i + 1 explicitly. But traceback from the first example (at least for 3.13) looks ok for me:

>>> def a():
...     i += 2
...     print(i)
...     
>>> a()
Traceback (most recent call last):
  File "<python-input-4>", line 1, in <module>
    a()
    ~^^
  File "<python-input-3>", line 2, in a
    i += 2
    ^
UnboundLocalError: cannot access local variable 'i' where it is not associated with a value

I don't think it's a bug.

[inhahe]

Why do you think it's a bug?

Documentation says: "If a name is bound in a block, it is a local variable of that block, unless declared as nonlocal or global."

In your last example you just do i = i + 1 explicitly. But traceback from the first example (at least for 3.13) looks ok for me:

>>> def a():
...     i += 2
...     print(i)
...     
>>> a()
Traceback (most recent call last):
  File "<python-input-4>", line 1, in <module>
    a()
    ~^^
  File "<python-input-3>", line 2, in a
    i += 2
    ^
UnboundLocalError: cannot access local variable 'i' where it is not associated with a value

I don't think it's a bug.

I realized it's not so much a bug but reporting the error in a weird place (for me) when using +=. It seems for you it reported it where the assignment is, for me it reported it later on. Not sure why it seems different for us. Oh, I see it did say line 2 for you, which is the same as it said for me (but it didn't underline the offending part), but in my actual program it reported the error way later after the assignment.

Here's my actual program:

import re
rules = []
ignore = ""
requotes = re.compile("""(["'])(.*)\\1""", re.S)
def parse(tokens, text):
  for line in tokens.split('\n'):
    try:
      token, rule = line.split('=', 1)
      token = token.strip()
      if token[0]=="#":
        continue
      ruleobj = requotes.match(rule.strip())
      # ruleobj = requotes.match[rule.strip()] for unicode
      rule = ruleobj.groups()[1]
      if token[0]=="_":
        ignore += rule 
      else:
        rules.append((rule, token))
    except:
      raise ValueError("Error in token input: " + line)

  if ignore:
    ignoreobj = re.compile(f"[{ignore}]")
  results = []
  curindex = 0
  for rule, token in rules:
    text = text[curindex:]
    matchobj = re.match(rule, text)
    if matchobj:
      results.append((token, text[:matchobj.end()], matchobj.groups()))
      curindex = matchobj.end()
      if ignore:
        matchobj = ignoreobj.match(text)
        if matchobj:
          curindex = matchobj.end()
      continue
  return results

Python showed an error in the if ignore line instead of the ignore += rule line.
(I did fix the bug in my program simply by moving ignore = "" to within the function.)

[skirpichev]

I see it did say line 2 for you, which is the same as it said for me

It's correct. Line 2 is not the print statement. It's augmented assignment. Don't forgot the function definition line (and lines counts from 1, not 0).

Here is an example on py3.8:

>>> i = 1
>>> def a():
...     print(1)
...     i += 1
... 
>>> a()
1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in a
UnboundLocalError: local variable 'i' referenced before assignment

[inhahe]

I see it did say line 2 for you, which is the same as it said for me

It's correct. Line 2 is not the print statement. It's augmented assignment. Don't forgot the function definition line (and lines counts from 1, not 0).

Here is an example on py3.8:

>>> i = 1
>>> def a():
...     print(1)
...     i += 1
... 
>>> a()
1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in a
UnboundLocalError: local variable 'i' referenced before assignment

Yeah, I see that it's correct in the interactive interpreter examples, but it didn't report it in the correct place in my actual program, perhaps because it's a string += instead of an int +=? That's the only meaningful difference I can think of. I included my actual program in my last reply to you (though maybe only after an edit, so maybe you didn't see it).

[skirpichev]

it didn't report it in the correct place in my actual program, perhaps because it's a string += instead of an int +=?

Please show how exactly you run your program. If I just run python3.13 a.py with your code snippet, I got nothing, of course.

[inhahe]

Oh, right.

>>> import mylexer
>>> print(mylexer.parse("""a = "[123]+?"\nb = '34'""", "341"))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "D:\temp\mylexer.py", line 22, in parse
    if ignore:
       ^^^^^^
UnboundLocalError: cannot access local variable 'ignore' where it is not associated with a value

[skirpichev]

Yes, you are correct, guessing that it's because you referenced local i before in code (with augmented assignment). Here is a simple example:

# a.py
i = 1
def f(a):
    if a:
        i += 1
    if i != 1:
        print(123)
f(0)

$ python3.13 -m dis a.py 
  0           RESUME                   0

  1           LOAD_CONST               0 (1)
              STORE_NAME               0 (i)

  2           LOAD_CONST               1 (<code object f at 0x7fd8d1a7ccf0, file "a.py", line 2>)
              MAKE_FUNCTION
              STORE_NAME               1 (f)

  7           LOAD_NAME                1 (f)
              PUSH_NULL
              LOAD_CONST               2 (0)
              CALL                     1
              POP_TOP
              RETURN_CONST             3 (None)

Disassembly of <code object f at 0x7fd8d1a7ccf0, file "a.py", line 2>:
  2           RESUME                   0

  3           LOAD_FAST                0 (a)
              TO_BOOL
              POP_JUMP_IF_FALSE        5 (to L1)

  4           LOAD_FAST_CHECK          1 (i)
              LOAD_CONST               1 (1)
              BINARY_OP               13 (+=)
              STORE_FAST               1 (i)

  5   L1:     LOAD_FAST_CHECK          1 (i)
              LOAD_CONST               1 (1)
              COMPARE_OP             119 (bool(!=))
              POP_JUMP_IF_FALSE       12 (to L2)

  6           LOAD_GLOBAL              1 (print + NULL)
              LOAD_CONST               2 (123)
              CALL                     1
              POP_TOP
              RETURN_CONST             0 (None)

  5   L2:     RETURN_CONST             0 (None)

Now with removed if block:

# b.py
$ python3.13 -m dis b.py 
  0           RESUME                   0

  1           LOAD_CONST               0 (1)
              STORE_NAME               0 (i)

  2           LOAD_CONST               1 (<code object f at 0x7f9afd27c930, file "b.py", line 2>)
              MAKE_FUNCTION
              STORE_NAME               1 (f)

  5           LOAD_NAME                1 (f)
              PUSH_NULL
              LOAD_CONST               2 (0)
              CALL                     1
              POP_TOP
              RETURN_CONST             3 (None)

Disassembly of <code object f at 0x7f9afd27c930, file "b.py", line 2>:
  2           RESUME                   0

  3           LOAD_GLOBAL              0 (i)
              LOAD_CONST               1 (1)
              COMPARE_OP             119 (bool(!=))
              POP_JUMP_IF_FALSE       12 (to L1)

  4           LOAD_GLOBAL              3 (print + NULL)
              LOAD_CONST               2 (123)
              CALL                     1
              POP_TOP
              RETURN_CONST             0 (None)

  3   L1:     RETURN_CONST             0 (None)

See https://docs.python.org/3/reference/executionmodel.html#resolution-of-names

[inhahe]

Yes, you are correct, guessing that it's because you referenced local i before in code (with augmented assignment). Here is a simple example:

Yeah, it's definitely because of the ignore += rule. I just think the error should have occurred at the assignment instead of at the if ignore: to make it more intuitive, and also more consistent i = i + 1 and i += 1 where i is an integer.

[skirpichev]

That's because in your case += operation had no chance to be executed. Try to change in my example f(0) to f(1).

[inhahe]

I think it was executed. One reason is that in debugging the problem I had commented out that line and I no longer got the error. Another reason is that if that line isn't executed then it's equivalent to

i = 1
def a():
   print(i)
a()

which is perfectly valid and doesn't render an error.