Scheme using and operator on reduce

Question

I am trying to use logical operations on reduce, for example:

(reduce and '(#t #t #t) 0)

This gives me an error, I'm not sure why.

(reduce + '(1 2 3) 0)

This works perfectly fine, but when I try to use the built-in and operation, it fails. Can someone explain me why wouldn't this work? I am forced to use reduce on logical operation

and is not a function, as Terje D said, but you can make your own to pass around: (foldl (lambda (x y) (and x y)) #t '(#t #t)) returns #t :-) — Brian Fearn
– Brian Fearn, Commented Apr 20, 2014 at 20:42

amalloy · Accepted Answer · 2014-04-20 21:08:59Z

4

reduce requires a function as its first argument, while and is a special form. Since special forms are not first class objects, they can not be used as arguments to functions (or be stored in variables).

An alternative to reduce in this case is

(every identity list-of-booleans)

while

(any identity list-of-booleans)

is a working alternative to (reduce or ...)

edited Apr 20, 2014 at 21:08

amalloy

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answered Apr 20, 2014 at 20:36

Terje D.

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3 Comments

R Sahu Over a year ago

Are every and any standard Scheme functions/constructs?

Sylwester Over a year ago

@RSahu Almost. Like reduce they are specified in SRFI-1 List library

user3100209 Over a year ago

@Terje D. I am required to use reduce function. I think I can work my way around since I know and is special form and not a operator. Making my own and would be easier I think.. Thank you very much

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