view tests/test-lrucachedict.py @ 33331:4bae3c117b57

scmutil: make cleanupnodes delete divergent bookmarks

cleanupnodes takes care of bookmark movement, and bookmark movement could
cause bookmark divergent resolution as a side effect. This patch adds such
bookmark divergent resolution logic so future rebase migration will be
easier.

The revset is carefully written to be equivalent to what rebase does today.
Although I think it might make sense to remove divergent bookmarks more
aggressively, for example:

    F   book@1
    |
    E   book@2
    |
    | D book
    | |
    | C
    |/
    B   book@3
    |
    A

When rebase -s C -d E, "book@1" will be removed, "book@3" will be kept,
and the end result is:

    D   book
    |
    C
    |
    F
    |
    E   book@2 (?)
    |
    B   book@3
    |
    A

The question is should we keep book@2? The current logic keeps it. If we
choose not to (makes some sense to me), the "deleterevs" revset could be
simplified to "newnode % oldnode".

For now, I just make it compatible with the existing behavior. If we want to
make the "deleterevs" revset simpler, we can always do it in the future.

from __future__ import absolute_import, print_function

from mercurial import (
    util,
)

def printifpresent(d, xs, name='d'):
    for x in xs:
        present = x in d
        print("'%s' in %s: %s" % (x, name, present))
        if present:
            print("%s['%s']: %s" % (name, x, d[x]))

def test_lrucachedict():
    d = util.lrucachedict(4)
    d['a'] = 'va'
    d['b'] = 'vb'
    d['c'] = 'vc'
    d['d'] = 'vd'

    # all of these should be present
    printifpresent(d, ['a', 'b', 'c', 'd'])

    # 'a' should be dropped because it was least recently used
    d['e'] = 've'
    printifpresent(d, ['a', 'b', 'c', 'd', 'e'])

    assert d.get('a') is None
    assert d.get('e') == 've'

    # touch entries in some order (get or set).
    d['e']
    d['c'] = 'vc2'
    d['d']
    d['b'] = 'vb2'

    # 'e' should be dropped now
    d['f'] = 'vf'
    printifpresent(d, ['b', 'c', 'd', 'e', 'f'])

    d.clear()
    printifpresent(d, ['b', 'c', 'd', 'e', 'f'])

    # Now test dicts that aren't full.
    d = util.lrucachedict(4)
    d['a'] = 1
    d['b'] = 2
    d['a']
    d['b']
    printifpresent(d, ['a', 'b'])

    # test copy method
    d = util.lrucachedict(4)
    d['a'] = 'va3'
    d['b'] = 'vb3'
    d['c'] = 'vc3'
    d['d'] = 'vd3'

    dc = d.copy()

    # all of these should be present
    print("\nAll of these should be present:")
    printifpresent(dc, ['a', 'b', 'c', 'd'], 'dc')

    # 'a' should be dropped because it was least recently used
    print("\nAll of these except 'a' should be present:")
    dc['e'] = 've3'
    printifpresent(dc, ['a', 'b', 'c', 'd', 'e'], 'dc')

    # contents and order of original dict should remain unchanged
    print("\nThese should be in reverse alphabetical order and read 'v?3':")
    dc['b'] = 'vb3_new'
    for k in list(iter(d)):
        print("d['%s']: %s" % (k, d[k]))

if __name__ == '__main__':
    test_lrucachedict()