view mercurial/dicthelpers.py @ 22377:f98abe3146b2 stable

dispatch: check shell alias again after loading extensions (issue4355)

Before this patch, the shell alias causes failure when it takes its
specific (= unknown for "hg") options in the command line, because
"_parse()" can't accept them.

This is the regression introduced by 03d345da0579.

It fixed the issue that ambiguity between shell aliases and commands
defined by extensions was ignored. But it also caused that ambiguous
shell alias is handled in "_parse()" even if it takes specific options
in the command line.

To avoid such failure, this patch checks shell alias again after
loading extensions.

All aliases and commands (including ones defined by extensions) are
completely defined before the 2nd (= newly added in this patch)
"_checkshellalias()" invocation, and "cmdutil.findcmd(strict=False)"
can detect ambiguity between them correctly.

For efficiency, this patch does:

  - omit the 2nd "_checkshellalias()" invocation if "[ui] strict= True"

    it causes "cmdutil.findcmd(strict=True)", of which result should
    be equal to one of the 1st invocation before adding aliases

  - avoid removing the 1st "_checkshellalias()" invocation

    it causes "cmdutil.findcmd(strict=True)" invocation preventing
    shell alias execution from loading extensions uselessly

# 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