[plt-scheme] Syntax question (from 4.2.1 of Report) -- never mind, I see it now

--- Gregory Woodhouse <gregory.woodhouse at sbcglobal.net> wrote:

> I'm puzzled by this example from the report
> 
> (cond ((assv 'b '((a 1) (b 2))) => cadr)
>        (else #f))
> 
> which evaluates to 2. I understand that (assv 'b '((a 1) (b 2)))  
> evaluates to (b 2), and cadr is an abbreviation for the car of the  
> cdr, but don't quite know what to make of the cond expression.
> 

Ah...never mind. I wasn't paying attention. I see now that if <test>
evaluates to true, then

(<test> => <expression>) applies <expression> to the value of <test>.
Since (assv 'b '((a 1) (b 2))) evaluates to (b 2), the result is just
(cadr '(b 2)) or 2.

I had just never seen the => notation before and mistook it for
something else.

===
Gregory Woodhouse  <gregory.woodhouse at sbcglobal.net>
"Interaction is the mind-body problem of computing."
--Philip L. Wadler