Op 2-4-2012 13:02, Stephen Bloch schreef:
On Apr 2, 2012, at 3:25 AM, Pierpaolo Bernardi quoted:
I thought that the position of the ufo can be find at (posn-y p) and
(posn-x) because the postition is a struct of Posn.
and replied:
You must pay more attention to small details. What is the p in (posn-y
p)? Isn't an argument missing in (posn-x)?
What is the u in (ufo-loc u) (that's the variable that DrRacket says
it doesn't know).
When I teach structs, I tell my students to write down the data definition (in a natural-language
comment), then a define-struct, then contracts for all the "functions that come for
free." For example, if I were defining "posn" from scratch, it would look like
; A posn has two numbers (x and y).
(define-struct posn [x y])
; make-posn : number(x) number(y) -> posn
; posn-x : posn -> number
; posn-y : posn -> number
; posn? : anything -> boolean
That's in the HtDP student languages, of course; in #lang racket the
constructor would be named posn rather than make-posn.
Once you've written these contracts, make sure you're actually following them.
"(posn-x)" doesn't follow its contract because posn-x is supposed to take a
parameter.
"p-x" doesn't follow the contract because it's not defined at all.
"(posn-y u)" doesn't follow the contract because u is a ufo, and posn-y is
supposed to take in a posn.
And so on.
Stephen Bloch
sbl...@adelphi.edu
Like you say U is the variable ufo.
I like the way you do these things.
but what did you mean with "p-x" ?
Following your way of working the answer is (posn-x p) and (posn-y p)
because P is the struct which holds the position.
Roelof
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