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.
