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Apply two optimizations to bitwise operations.

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Based on basic Boolean algebra: distributive and absorption laws.
This commit is contained in:
Sei Lisa 2017-10-12 22:48:43 +02:00
parent 9e7a5d1cdf
commit 7ae7fa4396
1 changed files with 38 additions and 1 deletions
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37
lslopt/lslfoldconst.py
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@ -122,6 +122,16 @@ class foldconst(object):
return True
return False
def CompareTrees(self, node1, node2):
"""Try to compare two subtrees to see if they are equivalent."""
# They MUSt be SEF
if 'SEF' not in node1 or 'SEF' not in node2:
return False
# So far it's only accepted if both are identifiers.
return (node1['nt'] == node2['nt'] == 'IDENT'
and node1['name'] == node2['name']
and node1['scope'] == node2['scope'])
def FoldStmt(self, parent, index):
"""Simplify a statement."""
node = parent[index]
@ -985,6 +995,33 @@ class foldconst(object):
# a&0 -> 0 if a is SEF
if 'SEF' in child[a]:
parent[index] = child[b]
# Apply boolean distributivity
opposite = '&' if nt == '|' else '|'
if child[0]['nt'] == child[1]['nt'] == opposite:
left = child[0]['ch']
right = child[1]['ch']
for c, d in ((0, 0), (0, 1), (1, 0), (1, 1)):
if self.CompareTrees(left[c], right[d]):
child[1]['nt'] = nt
nt = node['nt'] = opposite
opposite = child[1]['nt']
right[d] = left[1 - c]
child[0] = left[c]
break
# Apply absorption, possibly after distributivity
if child[0]['nt'] == opposite or child[1]['nt'] == opposite:
c = 0 if child[1]['nt'] == opposite else 1
for d in (0, 1):
if (self.CompareTrees(child[c], child[1 - c]['ch'][d])
and 'SEF' in child[1 - c]['ch'][1 - d]
):
node = parent[index] = child[c]
nt = node['nt']
child = node['ch'] if 'ch' in node else None
break
return
if nt == '^':