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schemeoutputcons

Scheme Cons (with #f as a second statement) output

I was wondering, If

(cons (quote (a b c)) #f)

gives an output

(( a  b  c ))

Then what output does this give:

(cons (quote (a b c)) #t)

?

Thank you

Solution

  • The first expression will not evaluate to ((a b c)) in most interpreters, it seems that your interpreter is evaluating #f as an empty list:

    (cons (quote (a b c)) '())
=> '((a b c))

    Having said that, you just substituted a #f with a #t, the standard results will look like this:

    (cons (quote (a b c)) #f)
=> '((a b c) . #f)

(cons (quote (a b c)) #t)
=> '((a b c) . #t)

    Why don't you try it online? in here for instance.