Scheme : using lambda as a parameter

Question

hy everyone, for school i have to make a function where lambda is used as a parameter

like so : (string (lambda ...) 5 40) where we have to fill in the dots

this is the function we had to reinvent, the regular string version

(define (string decoration n r)            >string decoration is a function that creates a string with either fish or pumpkins hanging on the string
  (define (decorations k)                  >decorations is the recursive function which hangs all the decorations together
    (if (= k 1)
        (decoration r 10)                  > here decoration is to be replaced with either a pumpkin or a fish as stated in the parameters
        (ht-append (decoration r 10)       > ht-append is a function that appends 2 figures Horizontally at the Top
                   (decorations (- k 1)))))
  (hang-by-thread (decorations n)))        > hang by thread is a function that hangs all the decorations at a string

all the names should be self-explanatory, the function takes a decoration , either a fish or a pumpkin and hangs it by a thread. But the fish has 3 parameters and the pumpkin has 2 which caused an error. So in a previous exercise we had to make an extra definition called fish-square which uses only 2 parameters to make a fish. Now we have to implement this same squared fish but with a lambda. Any help is greatly appreciated

(define (fish-square wh l)                        > this is the fish square functio which makes a fish but with 2 times the same parameter so it looks like a square
  (vc-append (filled-rectangle 2 l) (fish wh wh)))   > the l is the length of the string that attaches the fish to the string at the top

the fish function is just (fish x y) x makes it longer, y makes it taller. the pumpkin function is just (pumpkin x y) same story

so my question is, how do rewrite the given code , but with lambda as a parameter. i would upload an image, but my repuation isn't high enough :s

Re: "all the names should be self-explanatory" (i) You mentioned ht-append in your other question, as well. What is it? (ii) What does decoration do? Why does it take whatever r is and a number. (iii) If people haven't read your earlier question (or its duplicate, which you haven't deleted yet), the mention of fish and pumpkins isn't going to make much sense. — Joshua Taylor
– Joshua Taylor, Commented Oct 30, 2013 at 20:11
Also, the question is tagged with scheme, there's no need to put it in the title, too. — Joshua Taylor
– Joshua Taylor, Commented Oct 30, 2013 at 20:11

Óscar López · Accepted Answer · 2013-10-30 19:35:01Z

2

The string procedure as it is already receiving a procedure as a parameter (you don't have to rewrite it!), decoration can be any two-argument function used for decorating. Now when you call it you can pass a named procedure, for example:

(define (decoration r n)
  <body>)

(string decoration
        5
        40)

... Or just as easily, you can pass the same procedure in-line as a lambda, and if I understood correctly, this is what you're supposed to do:

(string (lambda (r n)
          <body>)
        5
        40)

Just replace <body> with the actual body of the decoration you want to use. In othre words: the change you're expected to do is in the way you pass the parameters to the function at invocation time, but you're not expected to change the function itself.

edited Oct 30, 2013 at 19:35

answered Oct 30, 2013 at 19:19

Óscar López

237k38 gold badges321 silver badges391 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sylwester · Accepted Answer · 2013-10-30 22:28:08Z

0

Imagine you have the procedure +. It could be any really. It takes several arguments but you need a different procedure that takes one and adds that to an already constant value 3.

Thus you want to pass + with the extra information that it should add 3.

A full definition of such procedure would be

(define (add3 n)
   (+ 3 n))

which is the short form of the full define

(define add3 
   (lambda (n)
      (+ 3 n)))

Now when passing a procedure 3+ you could actually just pass it's definition. These two does the same:

(do-computation add3 data)

(do-computation (lambda (n) (+ 3 n)) data)

answered Oct 30, 2013 at 22:28

Sylwester

48.9k4 gold badges52 silver badges83 bronze badges

Collectives™ on Stack Overflow

Scheme : using lambda as a parameter

2 Answers 2

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related