ast.parse creates only Constant(value=0) nodes for a `case +0:` pattern, which is inconsistent.

Bug report

Bug description:

I found the following asymmetry in the generated ast while I was working with the new cpython 3.15.

c="""
match 0:
    case +0:
        pass
    case -0:
        pass
+0
-0
"""

import ast

print(ast.dump(ast.parse(c),indent=2))

output (Python 3.15.0b3+dev):

Module(
  body=[
    Match(
      subject=Constant(value=0),
      cases=[
        match_case(
          pattern=MatchValue(
            value=Constant(value=0)),
          body=[
            Pass()]),
        match_case(
          pattern=MatchValue(
            value=UnaryOp(
              op=USub(),
              operand=Constant(value=0))),
          body=[
            Pass()])]),
    Expr(
      value=UnaryOp(
        op=UAdd(),
        operand=Constant(value=0))),
    Expr(
      value=UnaryOp(
        op=USub(),
        operand=Constant(value=0)))])

+0 is parsed to Constant(value=0) instead of UnaryOp(op=UAdd(),operand=Constant(value=0)) when it is used as a match-expression.

This was really unexpected for me and can maybe produce some problems with ast.parse/unparse roundtripps or code which works with the ast and require workarounds (like in my case)

Is it possible to implement this in a way that it generates UnaryOp(...)?

@johnslavik you implemented this #148566, maybe you can help me here.

CPython versions tested on:

3.15

Operating systems tested on:

No response

Linked PRs

• gh-152712

Beginne mit dem ast.parse-Beispiel im Issue und vergleiche die erzeugten Knoten für case +0:, case -0:, +0 und -0. Verfolge, wie der match-expression dargestellt wird, sorge dann dafür, dass das +0-Muster die erwartete UnaryOp-Form erzeugt, und überprüfe die AST-Ausgabe sowie den ast.parse/unparse-Roundtrip.