CS 536: Programming Language Concepts
Examples of CPS

To CPS-convert a cond statement:

• If the questions are all tail calls, send the answer of each cond-clause to the continuation k, then check whether the new answer expressions have any tail calls. Convert as needed.

• If a question has a non-tail call, pull the call outside the cond and replace the question with the box (the arg to the lambda). Then continue the conversion as before.

Examples

• (define (rect-area w h)           
    (* w h))
  (+ (rect-area 10 8) 3)
  
  ==>
  
  (define (rect-area/k w h k)
    (k (* w h)))
  
  (rect-area/k 10 8 (lambda (v) (+ v 3)))
  

• (define (g x) (* x 6))
  (define (h y) (+ y 5))
  (+ 4 (g (h 3)))
  
  [hint: convert to using h/k and g first, then to h/k and g/k]
  
  ==>
  
  (define (g/k x k)
    (k (* x 6)))
  
  (define (h/k y k)
    (k (+ y 5)))
  
  (h/k 3 (lambda (v1) (+ 4 (g v1))))
  
  ==>
  
  (h/k 3 (lambda (v1) (g/k v1 (lambda (v2) (+ 4 v2)))))
  

• (define (foo y) (+ (h (* y 2)) 7)) [h from above]
  (foo 8)
  
  ==>
  
  (define (foo/k y k) (k (+ (h (* y 2)) 7))) [h from above]
  
  ==>
  
  (define (foo/k y k) (h/k (* y 2) (lambda (v) (k (+ v 7)))))
  (foo/k 8 (lambda (x) x))
  

• (define (sum-area w1 h1 w2 h2)
    (+ (rect-area w1 h1)
       (rect-area w2 h2)))
  
  (sum-area 2 4 6 8)
  
  ==>
  
  (define (sum-area/k w1 h1 w2 h2 k)
    (rect-area/k w1 w2 (lambda (vleft) (k (+ vleft (rect-area w2 h2))))))
  
  ==>
  
  (define (sum-area/k w1 h1 w2 h2 k)
    (rect-area/k w1 h1 (lambda (vleft)
  		       (rect-area/k w2 h2 (lambda (vright)
  					    (k (+ vleft vright)))))))
  
  (sum-area 2 4 6 8 (lambda (x) x))
  
  [note that we have "linearized" the "tree-like" calls to rect-area]
  

• (define (product alon)
    (cond [(empty? alon) 1]
  	[(cons? alon) (* (first alon) (product (rest alon)))]))
  
  ==>
  
  [send both answers to k]
  
  (define (product/k alon k)
    (cond [(empty? alon) (k 1)]
  	[(cons? alon) (k (* (first alon) (product (rest alon))))]))
  
  
  ==>
  
  [convert the non-tail call in the cons? answer]
  
  (define (product/k alon k)
    (cond [(empty? alon) (k 1)]
  	[(cons? alon) (product/k (rest alon)
  				 (lambda (box)
  				   (k (* (first alon) box))))]))
  
  

• (define (filter pred L)
    (cond [(empty? L) empty]
  	[(cons? L)
  	 (cond [(pred (first L)) (cons (first L) (filter pred (rest L)))]
  	       [else (filter pred (rest L))])]))
  
  ==> 
  
  [send both answers to k]
  
  (define (filter/k pred L k)
    (cond [(empty? L) (k empty)]
  	[(cons? L)
  	 (k (cond [(pred (first L)) (cons (first L) (filter pred (rest L)))]
  		  [else (filter pred (rest L))])])))
  
  
  ==> 
  
  [pull up the non-tail call in the question (pred (first L))]
  
  (define (filter/k pred/k L k)
    (cond [(empty? L) (k empty)]
  	[(cons? L)
  	 (pred/k (first L)
  		 (lambda (predval)
  		   (k (cond [predval (cons (first L) (filter pred (rest L)))]
  			    [else (filter pred (rest L))]))))]))
  
  ==> 
  
  [send both answers of hte inner cond to k]
  
  (define (filter/k pred/k L k)
    (cond [(empty? L) (k empty)]
  	[(cons? L)
  	 (pred/k (first L)
  		 (lambda (predval)
  		   (cond [predval (k (cons (first L) (filter pred (rest L))))]
  			 [else (k (filter pred (rest L)))])))]))
  
  ==> 
  
  [work the k into both answer expressions]
  
  (define (filter/k pred/k L k)
    (cond [(empty? L) (k empty)]
  	[(cons? L)
  	 (pred/k (first L)
  		 (lambda (predval)
  		   (cond [predval
  			  (filter/k pred/k (rest L)
  				    (lambda (box) (k (cons (first L) box))))]
  			 [else (k (filter pred (rest L)))])))]))
  
  ==> 
  
  (define (filter/k pred/k L k)
    (cond [(empty? L) (k empty)]
  	[(cons? L)
  	 (pred/k (first L)
  		 (lambda (predval)
  		   (cond [predval
  			  (filter/k pred/k (rest L)
  				    (lambda (box) (k (cons (first L) box))))]
  			 [else (filter/k pred/k (rest L) k)])))]))