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Optimize a common condition of (x & flag1) && (x & flag2)

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It can't be done always: flag1 and flag2 must be nonzero powers of two. In that case, we can transform it to:

!~(x|~(flag1|flag2)) = !~(x|constant)

The -2147483648 case has trouble with the sign hack detector and I couldn't trigger it.
This commit is contained in:
Sei Lisa 2017-10-13 02:49:01 +02:00
parent 6ad2ef04d7
commit 3a849fe4b9
1 changed files with 36 additions and 0 deletions
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36
lslopt/lslfoldconst.py
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@ -242,6 +242,42 @@ class foldconst(object):
self.FoldCond(parent, index, ParentIsNegation)
return
# Specific optimization to catch a bitwise test appearing frequently.
# If b and c are nonzero constant powers of two:
# !(a & b) | !(a & c) -> ~(a|~(b|c))
# e.g. if (a & 4 && a & 8) -> if (!~(a|-13))
if (nt == '|'
and child[0]['nt'] == '!' and child[0]['ch'][0]['nt'] == '&'
and child[1]['nt'] == '!' and child[1]['ch'][0]['nt'] == '&'
):
and1 = child[0]['ch'][0]['ch']
and2 = child[1]['ch'][0]['ch']
a, b, c, d = 0, 1, 0, 1
if and1[b]['nt'] != 'CONST':
a, b = b, a
if and2[d]['nt'] != 'CONST':
c, d = d, c
if and1[b]['nt'] == and2[d]['nt'] == 'CONST':
val1 = and1[b]['value']
val2 = and2[d]['value']
if (val1 and val2
# power of 2
and ((val1 & (val1 - 1)) == 0 or val1 == -2147483648)
and ((val2 & (val2 - 1)) == 0 or val2 == -2147483648)
and self.CompareTrees(and1[a], and2[c])
):
# Check passed
child[0] = and1[a]
child[1] = and1[b]
child[1]['value'] = ~(val1 | val2)
parent[index] = {'nt':'~', 't':'integer', 'ch':[node]}
if 'SEF' in node:
parent[index]['SEF'] = True
self.FoldCond(parent, index, ParentIsNegation)
return
del val1, val2
del a, b, c, d, and1, and2
if nt in self.binary_ops and child[0]['t'] == child[1]['t'] == 'integer':
if nt == '!=':
if child[0]['nt'] == 'CONST' and child[0]['value'] == 1 \