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shelve: remove useless and incorrect code paths for file access...
shelve: remove useless and incorrect code paths for file access This patch removes code paths in "shelvedfile.opener()", because: - explicit "vfs.mkdir()" invocation is useless "vfs.__call__()" for modes other than "read" creates parent directory of target file automatically by "util.ensuredirs()". - mode checking in "except IOError" code path is useless ENOENT occurs only for "read" mode, because target file is created forcibly for other modes. - there is no explicit "return" statement in the code path for "except IOError" if "mode[0] in 'wa'" this is incorrect, because None may be returnd unexpectedly, even though it seems the EEXIST case in the directory creation race for ".hg/shelved" and is very rare. this directory creation race is also treated in "util.ensuredirs()".

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dicthelpers.py
55 lines | 1.6 KiB | text/x-python | PythonLexer
# dicthelpers.py - helper routines for Python dicts
#
# Copyright 2013 Facebook
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
def diff(d1, d2, default=None):
'''Return all key-value pairs that are different between d1 and d2.
This includes keys that are present in one dict but not the other, and
keys whose values are different. The return value is a dict with values
being pairs of values from d1 and d2 respectively, and missing values
treated as default, so if a value is missing from one dict and the same as
default in the other, it will not be returned.'''
res = {}
if d1 is d2:
# same dict, so diff is empty
return res
for k1, v1 in d1.iteritems():
v2 = d2.get(k1, default)
if v1 != v2:
res[k1] = (v1, v2)
for k2 in d2:
if k2 not in d1:
v2 = d2[k2]
if v2 != default:
res[k2] = (default, v2)
return res
def join(d1, d2, default=None):
'''Return all key-value pairs from both d1 and d2.
This is akin to an outer join in relational algebra. The return value is a
dict with values being pairs of values from d1 and d2 respectively, and
missing values represented as default.'''
res = {}
for k1, v1 in d1.iteritems():
if k1 in d2:
res[k1] = (v1, d2[k1])
else:
res[k1] = (v1, default)
if d1 is d2:
return res
for k2 in d2:
if k2 not in d1:
res[k2] = (default, d2[k2])
return res