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Runaway recursion with keyword arguments in 3.14 #137883
Description
Bug report
Bug description:
def f(a=0):
f(a=0)
f()This code causes an infinite loop (until it runs out of memory) in Python 3.14. It raises RecursionError in Python 3.13. There is no problem when calling the function without a keyword argument; in that case, it also raises RecursionError as expected.
CPython versions tested on:
3.13, 3.14, CPython main branch
Operating systems tested on:
macOS
Activity
StanFromIreland commentedon Aug 17, 2025
I think we can classify this as a crash, since there is no error, I had to kill it manually. WDYT @picnixz ?
StanFromIreland commentedon Aug 17, 2025
adqm commentedon Aug 17, 2025
This bisects to c13e7d9
StanFromIreland commentedon Aug 17, 2025
cc @markshannon
ZeroIntensity commentedon Aug 17, 2025
This looks pretty critical, considering it's in the core interpreter. We should deal with this before 3.14.0, or at least come to a decision before then.
I think that "bug" is fine. "crash" should be reserved for signals that come from the C code.
gh-137883: Check the recursion limit for specialized keyword argument…
pythongh-137883: Check the recursion limit for specialized keyword ar…
[3.14] pythongh-137883: Check the recursion limit for specialized key…
pythongh-137883: Check the recursion limit for specialized keyword ar…
[3.14] gh-137883: Check the recursion limit for specialized keyword a…
[3.14] pythongh-137883: Check the recursion limit for specialized key…