Mercurial > public > mercurial-scm > hg
view mercurial/dicthelpers.py @ 20398:2bc520bd0ce0
discovery: improve "note: unsynced remote changes!" warning
This note (which actually is a warning) frequently caused confusion.
"unsynced" is not a well established user-facing concept in Mercurial and the
message was not very specific or helpful.
Instead, show a messages like:
remote has heads on branch 'default' that are not known locally: 6c0482d977a3
and show it before aborting on "push creates new remote head". This will also
give more of a hint in the case where the branch has been closed remotely and
'hg heads' thus not would show any new heads after pulling.
A similar (but actually very different) message was addressed in 6b618aa08b6e.
author | Mads Kiilerich <madski@unity3d.com> |
---|---|
date | Thu, 06 Feb 2014 02:19:38 +0100 |
parents | ed46c2b98b0d |
children |
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# 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