(fix) return topN when query and current are both empty.

refs #14

rplugin/fruzzy_test.py

@@ -9,20 +9,20 @@
     from python3.fruzzy import scoreMatches
 
 
-def scoreMatchesProxy(q, c, limit, key=None, ispath=True):
+def scoreMatchesProxy(q, c, limit=10, current="", key=None, ispath=True):
     def idfn(x):
         return x
     if key is None:
         key = idfn
     if useNative:
         idxArr = scoreMatchesStr(q, [key(x) for x in c],
-                                 "", limit, ispath)
+                                 current, limit, ispath)
         results = []
         for i in idxArr:
             results.append((c[i[0]], i[1]))
         return results
     else:
-        return scoreMatches(q, c, "", limit, key, ispath)
+        return scoreMatches(q, c, current, limit, key, ispath)
 
 
 lines = []
@@ -129,3 +129,10 @@ def test_for_bug_08():
     results = list(scoreMatchesProxy("fmf", c, 10, ispath=True))
     assert results[0][0] == \
         'rplugin/python3/denite/filter/matcher/fruzzymatcher.py'
+
+def test_must_return_topN_when_query_and_current_are_empty():
+    c = ["/this/is/fileone.txt", "/that/was/FileOne.txt"]
+    results = list(scoreMatchesProxy("", c, 10, ispath=False))
+    assert len(results) == 2
+    assert results[0][0] == c[0]
+    assert results[1][0] == c[1]

rplugin/python3/fruzzy_mod.nim

@@ -300,15 +300,20 @@ iterator fuzzyMatches(query:string, candidates: openarray[string], current: stri
                 heap.push((i, rank))
                 if findFirstN and count == limit * 5:
                     break
-    else: # if blank string just take N items based on levenshtien (rev)
+    elif query == "" and current != "" and ispath: # if query is empty just take N items based on levenshtien (rev)
         for i, x in candidates:
             if current != x:
                 heap.push((i, 300 - current.editDistance(x)))
-    count = 0
-    while count < limit and heap.size > 0:
-        let item =  heap.pop
-        yield item
-        count.inc
+    else: # just return top N items from candidates as is
+        for j in 0 ..< min(limit, candidates.len):
+            yield (j, 0)
+
+    if heap.size > 0:
+        count = 0
+        while count < limit and heap.size > 0:
+            let item =  heap.pop
+            yield item
+            count.inc
 
 proc scoreMatchesStr(query: string, candidates: openarray[string], current: string, limit: int, ispath:bool=true): seq[tuple[i:int, r:int]] {.exportpy.} =
     result = newSeq[tuple[i:int, r:int]](limit)