run-tests: bump the default timeout on Windows to 4x the normal value
There are a ridiculous number of tests that timeout on Windows with the 360 sec
default (~60). And because of the bug where timed out tests still run to
completion before the results are thrown away[1], the timeout does nothing but
waste time, so there's no reason to try to find a lower value that still works.
For reference on my system:
# Ran 909 tests, 116 skipped, 119 failed.
python hash seed:
2052473208
real 151m44.322s
user 0m0.077s
sys 0m0.046s
[1] I thought that I wrote a bug for this, but search isn't finding it.
# pushkey.py - dispatching for pushing and pulling keys
#
# Copyright 2010 Olivia Mackall <olivia@selenic.com>
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
from __future__ import annotations
from . import (
bookmarks,
encoding,
obsolete,
phases,
)
def _nslist(repo):
n = {}
for k in _namespaces:
n[k] = b""
if not obsolete.isenabled(repo, obsolete.exchangeopt):
n.pop(b'obsolete')
return n
_namespaces = {
b"namespaces": (lambda *x: False, _nslist),
b"bookmarks": (bookmarks.pushbookmark, bookmarks.listbookmarks),
b"phases": (phases.pushphase, phases.listphases),
b"obsolete": (obsolete.pushmarker, obsolete.listmarkers),
}
def register(namespace, pushkey, listkeys):
_namespaces[namespace] = (pushkey, listkeys)
def _get(namespace):
return _namespaces.get(namespace, (lambda *x: False, lambda *x: {}))
def push(repo, namespace, key, old, new):
'''should succeed iff value was old'''
pk = _get(namespace)[0]
return pk(repo, key, old, new)
def list(repo, namespace):
'''return a dict'''
lk = _get(namespace)[1]
return lk(repo)
encode = encoding.fromlocal
decode = encoding.tolocal
def encodekeys(keys):
"""encode the content of a pushkey namespace for exchange over the wire"""
return b'\n'.join([b'%s\t%s' % (encode(k), encode(v)) for k, v in keys])
def decodekeys(data):
"""decode the content of a pushkey namespace from exchange over the wire"""
result = {}
for l in data.splitlines():
k, v = l.split(b'\t')
result[decode(k)] = decode(v)
return result