mercurial/dicthelpers.py

phase: better error message when --force is needed

When trying to turn a draft changeset into a secret changeset, I was
told:

  % hg phase -s .
  cannot move 1 changesets to a more permissive phase, use --force
  no phases changed

That message struck me as being backwards -- the secret phase feels
less permissive to me since it restricts the changesets from being
pushed.

We don't use the word "permissive" elsewhere, 'hg help phase' talks
about "lower phases" and "higher phases". I therefore reformulated the
error message to be

  cannot move 1 changesets to a higher phase, use --force

That is not perfect either, but more in line with the help text. An
alternative could be

  cannot move phase backwards for 1 changesets, use --force

which fits better with the help text for --force.

# 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