函数call / cc的函数参数是否可以等效地调用延续或不调用延续而返回？

Question

The Scheme Programming Language says

Scheme allows the continuation of any expression to be captured with the procedure call/cc . call/cc must be passed a procedure p of one argument. call/cc constructs a concrete representation of the current continuation and passes it to p . The continuation itself is represented by a procedure k . Each time k is applied to a value, it returns the value to the continuation of the call/cc application. This value becomes, in essence, the value of the application of call/cc . If p returns without invoking k , the value returned by the procedure becomes the value of the application of call/cc .

Are the two following ways of defining p equivalent, as far as being called by call/cc is concerned:

• p returns without invoking k ,
• p calls k with its otherwise return value?

I am not sure how call/cc is defined. Does call/cc ever directly call the continuation k , besides indirectly via p calling k ?

Is it perfectly fine that both call/cc and p don't invoke continuation k ?

Answer 1

Yes, (call/cc (lambda (k) 1)) <=> (call/cc (lambda (k) (k 1))) . You can prove this by using the continuation passing style transform.

The key part is the CPS form of call/cc is (lambda (k) (lambda (f) ((fk) k))) . And the CPS forms of the two functions are (lambda (c) (lambda (k) (c 1))) and (lambda (c) (lambda (k) (k 1))) . Substitute and simplify and both result in (lambda (k) (k 1)) .

I greatly prefer delimited continuations as they have:

(reset (1 + (shift (lambda (f) f)))) <=> (lambda (v) (+ 1 v))

This can also be proved algebraically.

Answer 2

To work out the details of the follow up question we can use the code scheme-cps-convert.rkt from http://matt.might.net/articles/cps-conversion/

> (T-c '(call/cc (λ (k) 1)) 'halt)
'((λ (f cc) (f (λ (x _) (cc x)) cc)) (λ (k $k2873) ($k2873 1)) halt)
> (T-c '(call/cc (λ (k) (k 1))) 'halt)
'((λ (f cc) (f (λ (x _) (cc x)) cc)) (λ (k $k2940) (k 1 $k2940)) halt)

reducing the first expression:

  ((λ (f cc) (f (λ (x _) (cc x)) cc)) (λ (k $k2873) ($k2873 1)) halt)
= ((λ (k $k2873) ($k2873 1)) (λ (x _) (halt x)) halt)
= (halt 1)

and the second

  ((λ (f cc) (f (λ (x _) (cc x)) cc)) (λ (k $k2940) (k 1 $k2940)) halt)
= ((λ (k $k2940) (k 1 $k2940)) (λ (x _) (halt x)) halt)
= ((λ (x _) (halt x)) 1 halt)
= (halt 1)

proves they are equal, so using k or not using k in that position makes no difference.