Hello schemers,
My understanding, after going through initial chapters of SICP, of
define was that it is lambda in disguise. However, this code piece
(trying the solution to ex. 3.7 ed.1) does not work as expected and I
had to replace the define with lambda.
(define (estimate-integral x1 x2 y1 y2 trials)
(define experiment
(p (real-random x1 x2) (real-random y1 y2)))
(* (rect-area x1 x2 y1 y2) (monte-carlo trials experiment)))
The trace of the procedure showed that the experiment was getting
evaluated before calling the monte-carlo. When I replaced the
experiment with this code, it worked as it passed the procedure
instead of the value of the procedure
(define (estimate-integral x1 x2 y1 y2 trials)
(let ((experiment
(lambda()
(p (real-random x1 x2) (real-random y1 y2)))))
(* (rect-area x1 x2 y1 y2) (monte-carlo trials experiment))))
If lambda and define are one and the same then how come the evaluation
points are different? Any explanation is really appreciated.
