view mercurial/dicthelpers.py @ 19966:7985e3469f58 stable

largefiles: systematic testing of merges to/from largefiles 427ce5633c1c fixed one problem with update and added a test case for it. The test coverage was thus insufficient before that. To make sure we have good test coverage in this area we add systematic testing of all cases of merges that may or may not change normal files to largefiles or vice versa. The tests shows some annoying extra merge prompts in some cases, but these prompts are hard to avoid and they are now "safe" - they do not leave the system in a confused inconsistent state.
author Mads Kiilerich <madski@unity3d.com>
date Mon, 28 Oct 2013 22:34:05 +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