For lazy compilation there is one generic lazy compilation builtin that
spills all registers on the stack and then triggers the compilation of
the called function. Some of these registers may contain references.
If a GC was triggered during lazy compilation, the GC would have to
know which spill slots on the stack contain references.

This CL adds a check to guarantee that no GC can be triggered during
lazy compilation. Thereby it is not necessary for the GC to know which
spill slots contain references.

If successful, lazy compilation indeed does not allocate on the heap
and therefore cannot trigger a GC. However, when compilation fails, an
error objects needs to be allocated and thrown. This allocation may
trigger a GC, but that's not a problem, because the reference
parameters which may get corrupted by the GC will not be used anyways,
because the called function will never get executed after the failed
compilation.

R=clemensb@chromium.org

Fixes: v8:11366
Change-Id: Ic526d169d4e80ba83f517970ff234e669f854331
Reviewed-on: https://chromium-review.googlesource.com/c/v8/v8/+/3599474Reviewed-by: Clemens Backes <clemensb@chromium.org>
Commit-Queue: Andreas Haas <ahaas@chromium.org>
Cr-Commit-Position: refs/heads/main@{#80187}