Posted by Martin on February 17, 2004 at 01:11:39:
The grammer says:
ProcedureExp ::= "(" "lambda" "(" ( Identifier )* ")" Expression ")"
That means there doesn't have to be any parameters to the lambda. What is the resulting type of something like the following?
(lambda () (+ y x))
is it just
t_i -> int
?
Is it t0->int