It was working fine, but the style wasn't exactly the way I like it, because
I'm unnecessarily picky. :)
Also, removed the file extension since we should probably learn from the
fact that it's already been rewritten once from one language to another.
Who knows, maybe it will be again someday.
...only when running under minimal/do, of course.
The tests in question mostly fail because they're testing particular
dependency-related behaviour, and minimal/do doesn't support dependencies,
so naturally it doesn't work.
If a checksummed target A used to exist but is now missing, and we tried to
redo-ifchange that exact file, we would unnecessarily run 'redo-oob A A';
that is, we have to build A in order to determine if A needs to be built.
The sub-targets of redo-oob aren't run with REDO_UNLOCKED, so this would
deadlock instantly.
Add an assertion to redo-oob to ensure we never try to redo-ifchange the
primary target (thus converting the deadlock into an exception). And skip
doing redo-oob when the target is already the same as the thing we have to
check.
We were giving up and rebuilding the toplevel object, which did eventually
rebuild our checksummed file, but then the file turned out to be identical
to what it was before, so that nobody *else* who depended on it ended up
getting rebuilt. So the results were indeterminate.
Now we treat it as if its dirtiness is unknown, so we build it using
redo-oob before building any of its dependencies.
If a depends on b depends on c, and c is dirty but b uses redo-stamp
checksums, then 'redo-ifchange a' is indeterminate: we won't know if we need
to run a.do unless we first build b, but the script that *normally* runs
'redo-ifchange b' is a.do, and we don't want to run that yet, because we
don't know for sure if b is dirty, and we shouldn't build a unless one of
its dependencies is dirty. Eek!
Luckily, there's a safe solution. If we *know* a is dirty - eg. because
a.do or one of its children has definitely changed - then we can just run
a.do immediately and there's no problem, even if b is indeterminate, because
we were going to run a.do anyhow.
If a's dependencies are *not* definitely dirty, and all we have is
indeterminate ones like b, then that means a's build process *hasn't
changed*, which means its tree of dependencies still includes b, which means
we can deduce that if we *did* run a.do, it would end up running b.do.
Since we know that anyhow, we can safely just run b.do, which will either
b.set_checked() or b.set_changed(). Once that's done, we can re-parse a's
dependencies and this time conclusively tell if it needs to be redone or
not. Even if it does, b is already up-to-date, so the 'redo-ifchange b'
line in a.do will be fast.
...now take all the above and do it recursively to handle nested
dependencies, etc, and you're done.
We were rebuilding the checksummed file every time because redo-ifchange was
incorrectly assuming that a child's changed_runid that's greater than my
changed_runid means I'm dirty. But if my checked_runid is >= the child's
checked_runid, then I'm clean, because my checksum didn't change.
Clear as mud?
A new redo-stamp program takes whatever you give it as stdin and uses it to
calculate a checksum for the current target. If that checksum is the same
as last time, then we consider the target to be unchanged, and we set
checked_runid and stamp, but leave changed_runid alone. That will make
future callers of redo-ifchange see this target as unmodified.
However, this is only "half" support because by the time we run the .do
script that calls redo-stamp, it's too late; the caller is a dependant of
the stamped program, which is already being rebuilt, even if redo-stamp
turns out to say that this target is unchanged.
The other half is coming up.
