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Fold ==0, ==1, ==-1 in general expressions, not just in conditions.

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Also add a couple FIXMEs.
This commit is contained in:
Sei Lisa 2017-10-17 02:38:08 +02:00
parent 82081b2206
commit a6a16c6e8f
1 changed files with 42 additions and 2 deletions
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44
lslopt/lslfoldconst.py
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@ -124,7 +124,10 @@ class foldconst(object):
def CompareTrees(self, node1, node2): def CompareTrees(self, node1, node2):
"""Try to compare two subtrees to see if they are equivalent.""" """Try to compare two subtrees to see if they are equivalent."""
# They MUSt be SEF # They MUST be SEF
# FIXME: They must also be stable, i.e. return the same value
# in two successive calls with a certain degree of certainty.
# Counterexamples are llFrand, llGetTimestamp.
if 'SEF' not in node1 or 'SEF' not in node2: if 'SEF' not in node1 or 'SEF' not in node2:
return False return False
# So far it's only accepted if both are identifiers or function calls, # So far it's only accepted if both are identifiers or function calls,
@ -985,6 +988,42 @@ class foldconst(object):
return return
return return
if nt == '==':
if child[0]['t'] == child[1]['t'] == 'integer':
# Deal with operands in any order
a, b = 0, 1
if child[b]['nt'] != 'CONST':
a, b = 1, 0
if child[b]['nt'] == 'CONST':
if child[b]['value'] in (-1, 0, 1):
node = child[a]
SEF = 'SEF' in node
if child[b]['value'] == -1:
node = {'nt':'~', 't':'integer', 'ch':[node]}
if SEF: node['SEF'] = True
elif child[b]['value'] == 1:
node = {'nt':'NEG', 't':'integer', 'ch':[node]}
if SEF: node['SEF'] = True
node = {'nt':'~', 't':'integer', 'ch':[node]}
if SEF: node['SEF'] = True
node = parent[index] = {'nt':'!', 't':'integer',
'ch':[node]}
if SEF: node['SEF'] = True
del child
self.FoldTree(parent, index)
return
# While this is tempting, it can only be done for identifiers.
# Counterexample: llFrand(1) == llFrand(1) would
# almost always return FALSE. After CompareTrees is fixed,
# we can reinstate it.
#if self.CompareTrees(child[0], child[1]):
# # a == a -> 1
# parent[index] = {'nt':'CONST', 't':'integer', 'value':1,
# 'SEF':True}
# return
return
if nt in ('<=', '>=') or nt == '!=' and child[0]['t'] != 'list': if nt in ('<=', '>=') or nt == '!=' and child[0]['t'] != 'list':
# Except for list != list, all these comparisons are compiled # Except for list != list, all these comparisons are compiled
# as !(a>b) etc. so we transform them here in order to reduce # as !(a>b) etc. so we transform them here in order to reduce
@ -997,7 +1036,8 @@ class foldconst(object):
if nt == '>': if nt == '>':
# Invert the inequalities to avoid doubling the cases to check. # Invert the inequalities to avoid doubling the cases to check.
# a>b --> b<a # a>b -> b<a
# FIXME: This is only possible if at most one is non-SEF.
nt = node['nt'] = '<' nt = node['nt'] = '<'
child[1], child[0] = child[0], child[1] child[1], child[0] = child[0], child[1]
# fall through to check for '<' # fall through to check for '<'