Taking the 'and' of a list by folding in Scheme

问题

In the book Structure and Interpretation of Computer Programs by H. Abelson and G. J. Sussman with J. Sussman, the accumulation or fold-right is introduced in Section 2.2.3 as follows:

(define (accumulate op initial sequence)
  (if (null? sequence)
      initial
      (op (car sequence)
          (accumulate op initial (cdr sequence)))))

I tried to use this to take the and of a list of Boolean variables, by writing:

(accumulate and 
            true 
            (list true true false))

However, this gave me the error and: bad syntax in DrRacket (with #lang sicp), and I had to do this instead:

(accumulate (lambda (x y) (and x y))
            true
            (list true true false))

Why? I believe it has something to do with how and is a special form, but I don't understand Scheme enough to say. Perhaps I'm just missing some obvious mistake...

回答1:

You answered your own question: and is a special form (not a normal procedure!) with special evaluation rules, and accumulate expects a normal procedure, so you need to wrap it inside a procedure.

To see why and is a special form, consider these examples that demonstrate that and requires special evaluation rules (unlike procedures), because it short-circuits whenever it finds a false value:

; division by zero never gets executed
(and #f (/ 1 0))
=> #f

; division by zero gets executed during procedure invocation
((lambda (x y) (and x y)) #f (/ 1 0))
=>  /: division by zero

来源：https://stackoverflow.com/questions/60092146/taking-the-and-of-a-list-by-folding-in-scheme