Fixes for unnecessary-dunder-call (PLC2801)

[VascoSch92]

The PR addresses issue #16053

The bugs addressed are:

• print((not 1).__add__(1))
• try: print(list("x".__add__(y for y in "y"))) except TypeError as e: print(e)
• print(("a" and "x").__contains__("x"))
• print(("" or "x").__contains__("x"))
• print(("" if False else "x").__contains__("x"))
• print((x := "x").__contains__("y"))
• print((not 0).__radd__(1))
• print(3 - (2).__add__(1))
• print("x".__add__(*"y"))

@MichaReiser I don't think I have the best implementation. Perhaps, you have some hints how to make the code better ;-)

[github-actions]

ruff-ecosystem results

Linter (stable)

✅ ecosystem check detected no linter changes.

Linter (preview)

✅ ecosystem check detected no linter changes.

[MichaReiser]

@VascoSch92 can you point towards the code that you think isn't ideal or can you explain me what you're trying to do?

[VascoSch92]

@MichaReiser Yes my bad.

The logic behind to find a fix for a dunder call of the form a.__dunder__(b) is the following:

1. Understand who has the precedenze between a and b, i.e., fix of the form a dunder_operation b or b dunder_operation a.

2. Understand if one or more of a, b or the whole expression has to be wrapped in parentheses, i.e., one of the following cases:

   • (a) dunder_operation b
   • a dunder_operation (b)
   • (a dunder_operation b)
   • (a dunder_operation (b))
   • and so on...

Therefore,

• I added a method called needs_parentheses (line 263), which determines whether an and b require parentheses. To achieve this, I simply check whether a or b contains a keyword. However, the method feels somewhat hardcoded, and there may be a more efficient way to determine if parentheses are needed.
• Note that I did not implement all possible ramifications of the logic described above. I only implemented the ones necessary to address the bugs reported in the related issue.