The INSERTIONSORT crash, all three rulings (FIX.org 2026-09-20):
- (bytes s) allocates a writable copy through the allocator surface —
context or (bytes s a), StorageExhausted with retry, a registry note in
dev builds (flan_bytes_dup, lowered like vec-new). (bytes-view s) is the
old zero-cost reinterpret, renamed, read-only by convention; every
in-repo reader swept over to it. (string b) unchanged.
- String constants were already read-only on both backends at -O0; now
pinned — bytes-copy.flan rows on LLVM/-O0/--x86, and dies_segv rows
asserting the write-through-view trap on both backends.
- A dev build installs a SIGSEGV/SIGBUS handler by the same dev-only
constructor slot that arms the registry: one line naming the address and
the innermost frame, then the trap-hook park — stopped, not dead, the
daemon serving. No agent: message and re-raise. Release builds untouched.
Pinned by trap_park over dev-segv.flan.
(Ptr Enemy) already says Enemy, at compile time, in the walk. What the renderer
lacked was any way to know whether the storage at the far end is still there —
and an allocation registry is exactly a record of which addresses it is still
true to read. So the inspector follows a live one and renders the pointee by the
same walk as anything else, and names what died at a dead one.
println does not, and the split is not squeamishness: spec-memory.md fixes what
a printed Ptr prints, a printed line belongs to the program and has to read the
same in a release build, and a release build has no registry to ask. The two
callers already differ in an emitter record; they differ in one more.
No address appears in the text. An address is not stable across two runs, so
printing one would make a rendering depend on where the heap landed — the rule
Render already follows for an allocator. What a reader wants from a dangling
pointer is what died.
registry.flan is one program read twice: a dev build answers for an address at
the heap, arena and pool tiers, and a release build answers 0 to all of it. The
arena row is the free-all Valgrind cannot see — this does not make memcheck
report it, it makes the same read answerable.
