backout: use commonancestorsheads for checking linear heritage
If two revisions are linearly related, there will only be one ancestor, and
commonancestors and commonancestorsheads would give the same result.
commonancestorsheads is however slightly simpler, faster and more correct.
[('string', 'string value'), ('bool1', 'true'), ('bool2', 'false'), ('boolinvalid', 'foo'), ('int1', '42'), ('int2', '-42'), ('intinvalid', 'foo')]
[('list1', 'foo'), ('list2', 'foo bar baz'), ('list3', 'alice, bob'), ('list4', 'foo bar baz alice, bob'), ('list5', 'abc d"ef"g "hij def"'), ('list6', '"hello world", "how are you?"'), ('list7', 'Do"Not"Separate'), ('list8', '"Do"Separate'), ('list9', '"Do\\"NotSeparate"'), ('list10', 'string "with extraneous" quotation mark"'), ('list11', 'x, y'), ('list12', '"x", "y"'), ('list13', '""" key = "x", "y" """'), ('list14', ',,,, '), ('list15', '" just with starting quotation'), ('list16', '"longer quotation" with "no ending quotation'), ('list17', 'this is \\" "not a quotation mark"'), ('list18', '\n \n\nding\ndong')]
---
'string value'
'true'
'false'
None
---
values.string is not a boolean ('string value')
True
False
False
False
True
---
42
-42
---
['foo']
['foo', 'bar', 'baz']
['alice', 'bob']
['foo', 'bar', 'baz', 'alice', 'bob']
['foo', 'bar', 'baz', 'alice', 'bob']
['abc', 'd"ef"g', 'hij def']
['hello world', 'how are you?']
['Do"Not"Separate']
['Do', 'Separate']
['Do"NotSeparate']
['string', 'with extraneous', 'quotation', 'mark"']
['x', 'y']
['x', 'y']
['', ' key = ', 'x"', 'y', '', '"']
[]
['"', 'just', 'with', 'starting', 'quotation']
['longer quotation', 'with', '"no', 'ending', 'quotation']
['this', 'is', '"', 'not a quotation mark']
['ding', 'dong']
[]
[]
['foo']
['foo']
['foo', 'bar']
['foo', 'bar']
['foo bar']
['foo', 'bar']
None
True
boolinvalid
intinvalid