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Big bunch of expressions optimized, boolean/bitwise, and some new ideas.

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a != -1 is now optimized to ~a in conditions. a == 1, a != 1, etc. are all optimized. a||b -> a|b is done. a^0, a^-1 are optimized. a&&b is transformed to !(!a|!b), but not yet into a&b when possible.
This commit is contained in:
Sei Lisa 2015-02-28 05:14:52 +01:00
parent 1dea1bd12c
commit 42ac090f03
1 changed files with 113 additions and 11 deletions
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124
lslopt/lslfoldconst.py
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@ -42,18 +42,64 @@ class foldconst(object):
if node['nt'] in ('V++', 'V--'):
node['nt'] = '++V' if node['nt'] == 'V++' else '--V';
def IsBool(self, node):
"""Some operators return 0 or 1, and that allows simplification of
boolean expressions. This function tells whether we know for sure
that the result is boolean."""
return False # TODO: implement IsBool
# Ideas include some functions like llSameGroup and llDetectedGroup.
def FoldCond(self, parent, index):
"""When we know that the parent is interested only in the truth value
of the node, we can perform further optimizations. This function deals
with them.
"""
node = parent[index]
if node['nt'] in ('CONST', 'IDENT', 'FLD'):
nt = node['nt']
if nt in ('CONST', 'IDENT', 'FLD'):
if node['nt'] == 'CONST':
node['t'] = 'integer'
node['value'] = -1 if lslfuncs.cond(node['value']) else 0
return # Nothing to do if it's already simplified.
# TODO: Implement FoldCond
child = node['ch'] if 'ch' in node else None
if nt == '!' and child[0]['nt'] == '!':
# bool(!!a) equals bool(a)
parent[index] = child[0]['ch'][0]
return
if nt == 'NEG':
# bool(-a) equals bool(a)
parent[index] = child[0]
return
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 \
or child[1]['nt'] == 'CONST' and child[1]['value'] == 1:
# a != 1 -> a - 1 (which FoldTree will transform to ~-a)
node['nt'] = '-'
else:
# This converts != to ^; FoldTree will simplify ^-1 to ~
# and optimize out ^0.
node['nt'] = '^'
self.FoldTree(parent, index)
elif nt == '==':
if child[0]['nt'] == 'CONST' and -1 <= child[0]['value'] <= 1 \
or child[1]['nt'] == 'CONST' and -1 <= child[1]['value'] <= 1:
# Transform a==b into !(a-b) if either a or b are in [-1, 1]
parent[index] = {'nt':'!', 't':'integer', 'ch':[node]}
node['nt'] = '-'
self.FoldTree(parent, index)
# TODO: on bitwise OR, detect if the operands are negated.
# If so treat it as an AND, checking if the operands are boolean
# to try to simplify them to an expression of the form !(a&b)
# when possible. Or if one is bool and the other is not, to an
# expression of the form !(a&-b) (if b is bool).
# TODO: on bitwise OR, detect if one operand is nonzero.
# In that case, if SEF, it can be simplified out.
def CopyNode(self, node):
# This is mainly for simple_expr so no need to go deeper than 1 level.
@ -128,7 +174,7 @@ class foldconst(object):
if nt == '!':
self.FoldTree(child, 0)
self.FoldCond(child, 0)
# !! does *not* cancel out, but !!! can be simplified to !
# !! does *not* cancel out (unless in cond), but !!! can be simplified to !
subexpr = child[0]
if 'SEF' in subexpr:
node['SEF'] = True
@ -426,18 +472,74 @@ class foldconst(object):
# have e.g. {<< {& x y} 3}; there will be explicit
# parentheses here always, so we don't need to worry.
# we have {<<, something, {CONST n}}, transform into {*, something, {CONST n}}
# we have {<<, something, {CONST n}}
# we transform it into {*, something, {CONST n}}
node['nt'] = '*'
child[1]['value'] = 1 << (child[1]['value'] & 31)
# Fall through to optimize product
else: # x << 0 --> x
parent[index] = child[0]
else:
pass # TODO: Eliminate redundancy (x*1, x*-1, x|0, x&-1, etc.)
# Include != to ^ and || to | and maybe && to &
# Note some cases e.g. x*0 can't be optimized away without side-effect analysis.
# But some cases like %1 can be turned into *0 to save bytes.
# Turn also % (power of 2) into & mask (oops, nope, negative doesn't work)
# Maybe turn != -1 into ~ in if()'s.
return
# TODO: Eliminate redundancy (x*1, x*-1, x|0, x&-1, etc.)
# Note some cases e.g. x*0 can't be optimized away without side-effect analysis.
# But some cases like %1 can be turned into &0 to save bytes.
# Turn also % (power of 2) into & mask (oops, nope, negative doesn't work)
# Other optimizations:
# x&0 -> 0 (if SEF)
# x|-1 -> -1 (if SEF)
# (a<=b) to !(a>b)
# (a>=b) to !(a<b)
# (a!=b) to !(a==b)
# !(a>const) to a<(const+1) if no overflow (4 variants)
# a>2147483647 to FALSE if SEF, otherwise convert to a&0
# a<-2147483648 to FALSE if SEF, otherwise convert to a&0
# a*2 => a+a
if nt == '^':
if child[0]['nt'] == 'CONST' and child[0]['value'] in (0, -1):
if child[0]['value'] == 0:
parent[index] = child[1]
else:
node['nt'] = '~'
node['ch'] = [child[1]]
elif child[1]['nt'] == 'CONST' and child[1]['value'] in (0, -1):
if child[1]['value'] == 0:
parent[index] = child[0]
else:
node['nt'] = '~'
node['ch'] = [child[0]]
return
if nt == '&&' or nt == '||':
SEF = 'SEF' in child[0] and 'SEF' in child[1]
if nt == '||':
parent[index] = node = {'nt':'!', 't':'integer', 'ch':[
{'nt':'!', 't':'integer', 'ch':[
{'nt':'|', 't':'integer', 'ch':[child[0], child[1]]}
]}]}
if SEF:
node['SEF'] = node['ch'][0]['SEF'] = node['ch'][0]['ch'][0]['SEF'] = True
else:
parent[index] = node = {'nt':'!', 't':'integer', 'ch':[
{'nt':'|', 't':'integer', 'ch':[
{'nt':'!', 't':'integer', 'ch':[child[0]]}
,
{'nt':'!', 't':'integer', 'ch':[child[1]]}
]}]}
if SEF:
node['SEF'] = node['ch'][0]['SEF'] = True
if 'SEF' in node['ch'][0]['ch'][0]['ch'][0]:
node['ch'][0]['ch'][0]['SEF'] = True
if 'SEF' in node['ch'][0]['ch'][1]['ch'][0]:
node['ch'][0]['ch'][1]['SEF'] = True
# Make another pass
self.FoldTree(parent, index)
return
return
if nt in self.assign_ops: