Don't copy defunct buffer contents when undoing OOK. #23968
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When undoing an OOK, `Perl_sv_backoff` will always copy the shifted buffer contents down to the start of the buffer. That's required when the contents are live, unfortunate when the contents are defunct but tiny, and can have a noticeably bad effect on performance when the contents are defunct but large. See Perl#23967 for an example of the latter. With this commit the copy will only occur if `SvOK(sv)`. This test is expected to indicate whether the buffer contents are active or defunct.
When a `my` SV goes out of scope, any OOK hack on its string buffer is undone by `Perl_sv_backoff`. If the SV is `SvOK`, a copy of the buffer contents will occur, but since the contents are defunct at this point, the copy is unnecessary. See Perl#23967 as an example of where this unnecessary copy had a noticeable effect on performance. This commit essentially inlines the necessary parts of `sv_backoff` to avoid the copy, without excessive messing around with `sv`'s flags at the call site in `Perl_leave_scope`.
richardleach
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tonycoz
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Dec 3, 2025
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Previously,
Perl_sv_backoffwould always copy the contents of the offsetstring buffer to the start of the buffer when undoing the OOK hack. When
the contents are defunct, this is unnecessary activity.
This PR makes two changes:
Perl_sv_backoffnow only copies whenSvOK(sv)is true.Perl_leave_scopefakes!SvOK(sv)when a lexical SV goes out of scope andsv_backoffis called.Closes GH #23967
Note:The main assumption made here is that
SvOK(sv)is a good proxy for buffer content liveness. It seems reasonable for core, but who knows what people have done in their own XS code.A lesser worry: Perhaps
Perl_leave_scopedoesn't need the SV flags preserving and it'd be okay to toggle theSvOKflags off and keep them off.