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RankedMatch: Do not compare word boundary match count on single word matches

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As the computation of word boundary matches is separate from the
actual subsequence matching, we sometimes have candidate that match
as a single word while still having multiple word boundary matches.

For example, with query "expresins", candidate "expressionism's"
will match as single word ("expressins" is a subsequence of
"expressionism"), and will have two word boundaries match (it does
match the last "s", which is considered as a separate word).

This should not be taken into account when compared against
candidate "expresions", which should be considered a better
match.

Fixes #1925
This commit is contained in:
Maxime Coste 2018-03-15 23:20:03 +11:00
parent 1ede2b89bb
commit 08df409a53
1 changed files with 4 additions and 1 deletions
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5
src/ranked_match.cc
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@ -180,7 +180,9 @@ bool RankedMatch::operator<(const RankedMatch& other) const
if (diff != Flags::None)
return (int)(m_flags & diff) > (int)(other.m_flags & diff);
if (not (m_flags & Flags::Prefix) and
// If we are SingleWord, FirstCharMatch will do the job, and we dont want to take
// other words boundaries into account.
if (not (m_flags & (Flags::Prefix | Flags::SingleWord)) and
m_word_boundary_match_count != other.m_word_boundary_match_count)
return m_word_boundary_match_count > other.m_word_boundary_match_count;
@ -241,6 +243,7 @@ UnitTest test_ranked_match{[] {
kak_assert(RankedMatch{"class", "cla"} < RankedMatch{"class::attr", "cla"});
kak_assert(RankedMatch{"meta/", "meta"} < RankedMatch{"meta-a/", "meta"});
kak_assert(RankedMatch{"find(1p)", "find"} < RankedMatch{"findfs(8)", "find"});
kak_assert(RankedMatch{"expresions", "expresins"} < RankedMatch{"expressionism's", "expresins"});
}};
UnitTest test_used_letters{[]()