Commit 7ca66772 authored by Tim Peters's avatar Tim Peters

get_matching_blocks(): rewrote code & comments so they match; added

more comments about why it's this way at all; and removed what looked
like needless expense (sorting (i, j, k) triples directly should give
exactly the same order as sorting (i, (i, j, k)) pairs).
parent 2adc626b
......@@ -474,30 +474,30 @@ class SequenceMatcher:
if self.matching_blocks is not None:
return self.matching_blocks
la, lb = len(self.a), len(self.b)
indexed_blocks = []
self.matching_blocks = matching_blocks = []
# This is most naturally expressed as a recursive algorithm, but
# at least one user bumped into extreme use cases that exceeded
# the recursion limit on their box. So, now we maintain a list
# ('queue`) of blocks we still need to look at, and append partial
# results to `matching_blocks` in a loop; the matches are sorted
# at the end.
queue = [(0, la, 0, lb)]
while queue:
# builds list of matching blocks covering a[alo:ahi] and
# b[blo:bhi], appending them in increasing order to answer
alo, ahi, blo, bhi = queue.pop()
i, j, k = x = self.find_longest_match(alo, ahi, blo, bhi)
# a[alo:i] vs b[blo:j] unknown
# a[i:i+k] same as b[j:j+k]
# a[i+k:ahi] vs b[j+k:bhi] unknown
i, j, k = x = self.find_longest_match(alo, ahi, blo, bhi)
if k:
if k: # if k is 0, there was no matching block
matching_blocks.append(x)
if alo < i and blo < j:
queue.append((alo, i, blo, j))
indexed_blocks.append((i, x))
if i+k < ahi and j+k < bhi:
queue.append((i+k, ahi, j+k, bhi))
indexed_blocks.sort()
self.matching_blocks = [elem[1] for elem in indexed_blocks]
self.matching_blocks.append( (la, lb, 0) )
return self.matching_blocks
matching_blocks.sort()
matching_blocks.append( (la, lb, 0) )
return matching_blocks
def get_opcodes(self):
"""Return list of 5-tuples describing how to turn a into b.
......
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