2010-11-12 22:47:03 -08:00
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import os
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2010-12-11 18:32:40 -08:00
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from atoi import atoi
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2010-11-12 22:47:03 -08:00
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2010-11-22 02:45:00 -08:00
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if not os.environ.get('REDO'):
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import sys
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sys.stderr.write('%s: error: must be run from inside a .do\n'
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% sys.argv[0])
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sys.exit(100)
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2010-11-21 03:34:32 -08:00
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PWD = os.environ.get('REDO_PWD', '')
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2010-11-13 00:11:34 -08:00
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TARGET = os.environ.get('REDO_TARGET', '')
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DEPTH = os.environ.get('REDO_DEPTH', '')
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2010-12-11 18:32:40 -08:00
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DEBUG = atoi(os.environ.get('REDO_DEBUG', ''))
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2010-11-21 06:23:41 -08:00
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DEBUG_LOCKS = os.environ.get('REDO_DEBUG_LOCKS', '') and 1 or 0
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2010-11-22 01:50:46 -08:00
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DEBUG_PIDS = os.environ.get('REDO_DEBUG_PIDS', '') and 1 or 0
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2010-11-13 00:11:34 -08:00
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VERBOSE = os.environ.get('REDO_VERBOSE', '') and 1 or 0
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2010-11-21 06:35:52 -08:00
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XTRACE = os.environ.get('REDO_XTRACE', '') and 1 or 0
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2010-11-21 07:10:48 -08:00
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KEEP_GOING = os.environ.get('REDO_KEEP_GOING', '') and 1 or 0
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redo-log: capture and linearize the output of redo builds.
redo now saves the stderr from every .do script, for every target, into
a file in the .redo directory. That means you can look up the logs
from the most recent build of any target using the new redo-log
command, for example:
redo-log -r all
The default is to show logs non-recursively, that is, it'll show when a
target does redo-ifchange on another target, but it won't recurse into
the logs for the latter target. With -r (recursive), it does. With -u
(unchanged), it does even if redo-ifchange discovered that the target
was already up-to-date; in that case, it prints the logs of the *most
recent* time the target was generated.
With --no-details, redo-log will show only the 'redo' lines, not the
other log messages. For very noisy build systems (like recursing into
a 'make' instance) this can be helpful to get an overview of what
happened, without all the cruft.
You can use the -f (follow) option like tail -f, to follow a build
that's currently in progress until it finishes. redo itself spins up a
copy of redo-log -r -f while it runs, so you can see what's going on.
Still broken in this version:
- No man page or new tests yet.
- ANSI colors don't yet work (unless you use --raw-logs, which gives
the old-style behaviour).
- You can't redirect the output of a sub-redo to a file or a
pipe right now, because redo-log is eating it.
- The regex for matching 'redo' lines in the log is very gross.
Instead, we should put the raw log files in a more machine-parseable
format, and redo-log should turn that into human-readable format.
- redo-log tries to "linearize" the logs, which makes them
comprehensible even for a large parallel build. It recursively shows
log messages for each target in depth-first tree order (by tracing
into a new target every time it sees a 'redo' line). This works
really well, but in some specific cases, the "topmost" redo instance
can get stuck waiting for a jwack token, which makes it look like the
whole build has stalled, when really redo-log is just waiting a long
time for a particular subprocess to be able to continue. We'll need to
add a specific workaround for that.
2018-11-03 22:09:18 -04:00
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RAW_LOGS = os.environ.get('REDO_RAW_LOGS', '') and 1 or 0
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2010-11-16 00:14:57 -08:00
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SHUFFLE = os.environ.get('REDO_SHUFFLE', '') and 1 or 0
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2010-12-11 19:08:53 -08:00
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STARTDIR = os.environ.get('REDO_STARTDIR', '')
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2010-12-11 18:32:40 -08:00
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RUNID = atoi(os.environ.get('REDO_RUNID')) or None
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2010-11-21 03:09:21 -08:00
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BASE = os.environ['REDO_BASE']
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while BASE and BASE.endswith('/'):
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BASE = BASE[:-1]
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The second half of redo-stamp: out-of-order building.
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.
2010-12-11 04:40:05 -08:00
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UNLOCKED = os.environ.get('REDO_UNLOCKED', '') and 1 or 0
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os.environ['REDO_UNLOCKED'] = '' # not inheritable by subprocesses
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2010-12-11 05:50:29 -08:00
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NO_OOB = os.environ.get('REDO_NO_OOB', '') and 1 or 0
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os.environ['REDO_NO_OOB'] = '' # not inheritable by subprocesses
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2016-11-27 23:35:28 -08:00
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def get_locks():
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"""Get the list of held locks."""
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return os.environ.get('REDO_LOCKS', '').split(':')
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def add_lock(name):
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"""Add a lock to the list of held locks."""
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locks = set(get_locks())
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locks.add(name)
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os.environ['REDO_LOCKS'] = ':'.join(list(locks))
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