##// END OF EJS Templates
test-pathencode: randomize length of each path component...
test-pathencode: randomize length of each path component This makes it possible for long and short components to exist in the same path. This also makes shorter path components more likely. For randint(1, randint(1, n)), the likelihood that one sees a number k (1 <= k <= n) is 1/n * (\sum_{k=i}^n 1/i). This decreases with k, much like in the real world where shorter paths are more common than longer ones. The previous fix and this one together cause issue3958 to be detected by this test with reasonable frequency. When this test was run 100 times in a loop, the issue was detected 30 of those times.

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pushkey.py
39 lines | 1.1 KiB | text/x-python | PythonLexer
# pushkey.py - dispatching for pushing and pulling keys
#
# Copyright 2010 Matt Mackall <mpm@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.
import bookmarks, phases, obsolete
def _nslist(repo):
n = {}
for k in _namespaces:
n[k] = ""
if not obsolete._enabled:
n.pop('obsolete')
return n
_namespaces = {"namespaces": (lambda *x: False, _nslist),
"bookmarks": (bookmarks.pushbookmark, bookmarks.listbookmarks),
"phases": (phases.pushphase, phases.listphases),
"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)