Using AND with the apply function in Scheme

Question

Why doesn't the following work?

(apply and (list #t #t #f))

While the following works just fine.

(apply + (list 1 3 2))

This seems to be the case in both R5RS and R6RS?

Answer 1

and isn't a normal function because it will only evaluate as few arguments as it needs, to know whether the result is true or false. For example, if the first argument is false, then no matter what the other arguments are, the result has to be false so it won't evaluate the other arguments. If and were a normal function, all of its arguments would be evaluated first, so and was made a special keyword which is why it cannot be passed as a variable.

Answer 2

(define and-l (lambda x 
    (if (null? x)
        #t
        (if (car x) (apply and-l (cdr x)) #f))))

pleas notice that this is lambda variadic! apply example (and-l #t #t #f)

or you can use it via apply procedure(as was asked) for example (apply and-l (list #t #t #f))

both options are ok...

Answer 3

and is actually a macro, whose definition is outlined in R5RS chapter 4. The notation "library syntax" on that page really means it is implemented as a macro.

Section 7.3, Derived expression types gives a possible definition of the and macro:

(define-syntax and
  (syntax-rules ()
    ((and) #t)
    ((and test) test)
    ((and test1 test2 ...)
     (if test1 (and test2 ...) #f))))

Given this defintion, it is not possible to use and as a function argument to apply.