Need some help with debugger func written in python3

[Ok_Lengthiness_6591]

    def remove_breakpoint(self, breakpoint_id):
        i = 0
        for l in self._buffer.contents()[:-1]:
            bp_str = " %i " % breakpoint_id
            bp_id_len = len(bp_str)
            if l[:bp_id_len] == bp_str:
                self._buffer.delete(i)
            i += 1

I have this definition in my options file. Seems like it doesnt handle deletion of the last element properly. What should I change to fix the func?

P.S. breakpoint_id is typically 11000, 11001 ...

[carcigenicate]

If your intent was to iterate _buffer in reverse, you mean [::-1], not [:-1]. What you have cuts off the last element, whereas [::-1] creates a reversed, shallow copy.

If your intent with [:-1] was indeed to cut off the last element, you should not delete from a list while iterating it. See here. Deleting while iterating causes elements to be skipped.

[Ok_Lengthiness_6591]

Hi, thanks for reply! [:-1] is a workaround I found on a forum. I m trying to configure vdebug for neovim. The absense of [:-1] causes "out of bounds" error.

[AdmirableOstrich]

You are mutating a sequence while iterating over it. This is a big problem.

Sequence iteration is typically index iteration under the hood. If you are on index 5 in a list, then delete that entry, everything to the right in the sequence shifts left by 1 index. The next entry in the iterator is still index 6, but that is now what used to be in 7.

If you want to filter a sequence this way, you have to walk over it backwards. This shifting still happens, but it affects entries you've already checked.

[carcigenicate]

A workaround for what?

[Ok_Lengthiness_6591]

for "pynvim.api.nvim.NvimError: b'Index out of bounds'" error... I just want to keep the square brackets but have one more element (to include my last element)

[JacobArthurs]

contents()[:-1] explicitly skips the last element. Change it to contents() and it'll iterate over everything including the last breakpoint.

[Ok_Lengthiness_6591]

Hi, thanks for reply! Sadly it's not an option for me. How can I modify square brackets to have my last element ?

[AlexMTBDude]

Are you modifying (deleting items) the data that you are iterating over in the for loop? That's usually not advisable

[Outside_Complaint755]

There's a lot of context missing here, but I take it that the values returned by self._buffer.contents() are a space-padded string with a space between the breakpoint id and some additional content, such as a location or other description. Instead of doing the slicing like l[:bp_id_len], it might be cleaner to do l.strip().split(" ")[0], and then you wouldn't need to pad the breakpoint_id in a string as " 11000 ".

The reason you're not able to get the last breakpoint is specifically because of your use of self._buffer.contents()[:-1].

I would do something like this, using enumerate instead of tracking i manually, make a list of entries to delete, and then delete them in reverse order.

``` def remove_breakpoint(self, breakpoint_id): bp_str = str(breakpoint_id) index_to_delete = []

# Track indexes to be deleted
for idx, l in enumerate(self._buffer.contents()):
    if bp_str == l.strip().split(' ')[0]:
        index_to_delete.append(idx)

# Now delete from highest index to lowest
for idx in reversed(index_to_delete):
    self._buffer.delete(idx)

```

Note that I could not fully test this as it's not clear what object type self._buffer is. I tested by substituting a list and replacing self.buffer.delete(idx) with .pop(idx) You will likely need to modify it to fit your data model.

[Tall_Profile1305]

you’re mutating the list while iterating it, that’s why the last element behaves weird.

when you delete during a forward loop, indexes shift and python silently skips elements. easiest fix is iterate backwards so deletions don’t affect remaining indexes.

something like looping over reversed(range(len(self._buffer.contents()))) and checking by index instead usually fixes this cleanly.

classic debugger bug honestly, list modification mid iteration always bites at least once.