def ReadWord(self, unused_lex_mode):
# type: (lex_mode_t) -> word__String
"""Interface for bool_parse.py.
TODO: This should probably be word_t
"""
if self.i == self.n:
# Does it make sense to define Eof_Argv or something?
# TODO: Add a way to show this location. Show 1 char past the right-most
# spid of the last word? But we only have the left-most spid.
w = word.String(Id.Eof_Real, '', runtime.NO_SPID)
return w
#log('ARGV %s i %d', self.argv, self.i)
s = self.cmd_val.argv[self.i]
left_spid = self.cmd_val.arg_spids[self.i]
self.i += 1
# default is an operand word
id_ = match.BracketUnary(s)
if id_ == Id.Undefined_Tok:
id_ = match.BracketBinary(s)
if id_ == Id.Undefined_Tok:
id_ = match.BracketOther(s)
if id_ == Id.Undefined_Tok:
id_ = Id.Word_Compound
# NOTE: We only have the left spid now. It might be useful to add the
# right one.
w = word.String(id_, s, left_spid)
return w
def _ThreeArgs(w_parser):
# type: (_StringWordEmitter) -> bool_expr_t
"""Returns an expression tree to be evaluated."""
w0 = w_parser.Read()
w1 = w_parser.Read()
w2 = w_parser.Read()
# NOTE: Order is important here.
binary_id = match.BracketBinary(w1.s)
if binary_id != Id.Undefined_Tok:
return bool_expr.Binary(binary_id, w0, w2)
if w1.s == '-a':
return bool_expr.LogicalAnd(bool_expr.WordTest(w0),
bool_expr.WordTest(w2))
if w1.s == '-o':
return bool_expr.LogicalOr(bool_expr.WordTest(w0),
bool_expr.WordTest(w2))
if w0.s == '!':
w_parser.Rewind(2)
child = _TwoArgs(w_parser)
return bool_expr.LogicalNot(child)
if w0.s == '(' and w2.s == ')':
return bool_expr.WordTest(w1)
p_die('Expected binary operator, got %r (3 args)', w1.s, word=w1)