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Fix corner case with labels.

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When 'if', 'for', and 'while' statements have a label as the sub-statement, e.g. `while (FALSE) @label;`, the label was improperly removed in some cases, causing a potential compilation failure.

Why someone would want to do that, one never knows, but just to be sure, it has been fixed.
This commit is contained in:
Sei Lisa 2017-01-05 02:55:06 +01:00
parent c7e8c04349
commit 63e4425cd5
1 changed files with 62 additions and 6 deletions
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68
lslopt/lslfoldconst.py
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@ -30,6 +30,7 @@ class foldconst(object):
scope = node['scope']
return self.symtab[scope][name]['Kind'] == 'v' \
and 'Loc' not in self.symtab[scope][name]
def FoldAndRemoveEmptyStmts(self, lst):
"""Utility function for elimination of useless expressions in FOR"""
idx = 0
@ -42,6 +43,18 @@ class foldconst(object):
else:
idx += 1
def DoesSomething(self, node):
"""Tell if a subtree does something or is just empty statements
(a pure combination of ';' and '{}')
"""
if node['nt'] != ';':
if node['nt'] == '{}':
if self.does_something(node['ch']):
return True
else:
return True
return False
def FoldStmt(self, parent, index):
"""Simplify a statement."""
node = parent[index]
@ -1095,20 +1108,38 @@ class foldconst(object):
# result of optimization so they must be wrapped in an
# IF statement). The current approach leaves unnecessary
# IFs behind.
if len(child) == 3:
if len(child) == 3 and child[2]['nt'] != '@':
del child[2] # Delete ELSE if present
return
else:
self.FoldStmt(child, 1)
if len(child) == 3 and child[2]['nt'] == '@':
# Corner case. The label is in the same scope as
# this statement, so it must be preserved just in
# case it's jumped to.
return
parent[index] = child[1]
return
elif len(child) == 3:
self.FoldTree(child, 2)
self.FoldStmt(child, 2)
if child[1]['nt'] == '@':
# Corner case. The label is in the same scope as this
# statement, so it must be preserved just in case it's
# jumped to.
if not self.DoesSomething(child[2]):
del child[2]
return
parent[index] = child[2]
return
else:
# No ELSE branch, replace the statement with an empty one.
if child[1]['nt'] == '@':
# Corner case. The label is in the same scope as this
# statement, so it must be preserved just in case it's
# jumped to.
parent[index] = child[1]
return
parent[index] = {'nt':';', 't':None, 'SEF':True}
return
else:
@ -1117,8 +1148,7 @@ class foldconst(object):
if len(child) > 2:
self.FoldTree(child, 2)
self.FoldStmt(child, 2)
if child[2]['nt'] == ';' \
or child[2]['nt'] == '{}' and not child[2]['ch']:
if not self.DoesSomething(child[2]):
# no point in "... else ;" - remove else branch
del child[2]
if all('SEF' in subnode for subnode in child):
@ -1141,8 +1171,14 @@ class foldconst(object):
self.FoldTree(child, 1)
self.FoldStmt(child, 1)
else:
# Can be removed.
parent[index] = {'nt':';', 't':None, 'SEF':True}
if child[1]['nt'] == '@':
# Corner case. The label is in the same scope as this
# statement, so it must be preserved just in case it's
# jumped to.
parent[index] = child[1]
else:
# Whole statement can be removed.
parent[index] = {'nt':';', 't':None, 'SEF':True}
return
else:
self.FoldTree(child, 1)
@ -1186,6 +1222,14 @@ class foldconst(object):
for expr in child[0]['ch']:
# Fold into expression statements.
exprlist.append({'nt':'EXPR', 't':expr['t'], 'ch':[expr]})
if (exprlist or child[2]['ch']) and child[3]['nt'] == '@':
# Corner case. We can't optimize this to one single
# statement, so we leave it as-is.
self.FoldTree(child, 3)
self.FoldStmt(child, 3)
self.FoldAndRemoveEmptyStmts(child[2]['ch'])
return
# returns type None, as FOR does
if exprlist:
# We're in the case where there are expressions. If any
@ -1193,7 +1237,19 @@ class foldconst(object):
# removed earlier) so don't mark this node as SEF.
parent[index] = {'nt':'{}', 't':None, 'ch':exprlist}
else:
parent[index] = {'nt':';', 't':None, 'SEF': True}
if child[3]['nt'] == '@':
# Corner case. The label is in the same scope as
# this statement, so it must be preserved. Also,
# jumping inside the loop would execute the
# iterator, so we fold it.
self.FoldAndRemoveEmptyStmts(child[2]['ch'])
if not child[2]['ch']:
# if there's something in the 2nd list,
# preserve the whole statement, otherwise
# replace it with the label
parent[index] = child[3]
else:
parent[index] = {'nt':';', 't':None, 'SEF': True}
return
else:
self.FoldTree(child, 3)