1. CS321 Functional Programming 2 © JAS 2005 6-1 The SECD Machine The SECD machine has become a classic way of implementing functional languages. It is based on an automaton designed by Peter Landin in the early sixties. As with the Combinator approach it assumes that the program is expressed as a λ-expression which is one of A variable An abstraction (bound variable part and body) An application (operator and operand)

2. CS321 Functional Programming 2 © JAS 2005 6-2 The SECD machine uses four stacks. These are The Stack which is used to store the partial results of evaluating an expression, and which will hold the final result of the machine's execution. The Environment which consists of a series of pairs that give values to the free variables of an expression; the value of a constant in any environment being that constant. The first element of any pair is an identifier and the second element is the associated value. The Control which stores the expression that is being evaluated. The Dump which is used to store copies of the four stacks whilst a sub-function is being evaluated.

3. CS321 Functional Programming 2 © JAS 2005 6-3 We can now present an outline algorithm for the implementation of the SECD machine. Since the design was intended to be used for implementation on a conventional sequential machine it is appropriate to present the algorithm in a sequential style.

4. CS321 Functional Programming 2 © JAS 2005 6-4 WHILE there is an expression to be evaluated OR there is a suspended computation to be resumed DO IF the evaluation of the current expression is complete THEN resume the last suspended computation with the newly computed value on S ELSE C is not empty D is not empty C is empty remember the value on S; recover S,E,C,D from D; push the value from the old S onto S

5. CS321 Functional Programming 2 © JAS 2005 6-5 CASE the next expression to be evaluated OFidentifier: push value of this identifier onto S and pop C λ-exp: push the appropriate closure onto S and pop C top of C find value of identifier in E; push it onto S; pop C push the triple (body, bvpart, E) onto S; pop C

6. CS321 Functional Programming 2 © JAS 2005 6-6 application: replace top of C by the expressions representing this application pop C; push the ap operator onto C; push the operator to be applied onto C; push the operand onto C

7. CS321 Functional Programming 2 © JAS 2005 6-7 ap: cause the operator on S to be applied to the operand below it IF hd(S) is a closure THEN push 4-tuple (tail(tail(S)),E,(tail(C),D) onto D; push body of closure onto new empty C; make E the environment from closure prefixed by pair associating bvpart of closure with hd(tail(S)); create a new empty S ELSE { the operator is a primitive function } pop C; find result of applying hd(S) to hd(tail(S)); pop S twice; push result onto S ENDC

8. CS321 Functional Programming 2 © JAS 2005 6-8 An example evaluation – (λx.λy.+xy) 3 4 SECD (λx.λy.+xy) 34 ap

9. CS321 Functional Programming 2 © JAS 2005 6-9 An example evaluation – (λx.λy.+xy) 3 4 SECD 4 ap (λx.λy.+xy)3 ap cl(λy.+xy,x,())

10. CS321 Functional Programming 2 © JAS 2005 6-10 An example evaluation – (λx.λy.+xy) 3 4 SECD 4 ap 3 cl(λy.+xy,x,()) (,(),,()) λy.+xy (x,3)

11. CS321 Functional Programming 2 © JAS 2005 6-11 An example evaluation – (λx.λy.+xy) 3 4 SECD 4 ap (,(),,()) λy.+xy (x,3) cl(+xy,y,(x,3))

12. CS321 Functional Programming 2 © JAS 2005 6-12 An example evaluation – (λx.λy.+xy) 3 4 SECD 4 ap (x,3) cl(+xy,y,(x,3)) (y,4) + x y ((),(),(),())

13. CS321 Functional Programming 2 © JAS 2005 6-13 An example evaluation – (λx.λy.+xy) 3 4 SECD (x,3) (y,4) ((),(),(),()) +xy ap 4 3

14. CS321 Functional Programming 2 © JAS 2005 6-14 An example evaluation – (λx.λy.+xy) 3 4 SECD (x,3) (y,4) ((),(),(),()) ap 4 3 + 7

15. CS321 Functional Programming 2 © JAS 2005 6-15 The SECD machine implements eager (applicative order) evaluation since it evaluates the arguments first. The evluation of the body of the function is delayed by the closure mechanism. A lazy SECD machine can be defined by introducing the concept of a suspension. Primitive functions may be regarded as strict, but other functions are lazy so their parameters are saved as a suspension which consists of the unevaluated parameter and the environment in which it is to be evaluated.

16. CS321 Functional Programming 2 © JAS 2005 6-16 The evaluation algorithm needs to add an extra case for a suspended computation, and also modify the actions required for an application and an ap operator. We will consider first the modification for the application case: application: replace top of C by the expressions representing this application pop C; push the ap operator onto C; push the operator to be applied onto C; push the operand onto C IF operand is function THEN push operand onto C ELSE create suspension – push pair (operand, E) onto S

17. CS321 Functional Programming 2 © JAS 2005 6-17 When evaluating the ap operator we have another option to consider:- If the operator is a closure, or has its necessary operands then the actions are as before. If the operator is a strict primitive function and its operand is a suspension we need to create a new set of stacks to evaluate the suspension. push the 4-tuple (S,E,C,D) onto D; push the sp operator onto C; push the body of the suspension onto C; push the environment of the suspension onto E; create a new empty S

18. CS321 Functional Programming 2 © JAS 2005 6-18 Finally we need to be able to evaluate the sp operator when it is at the top of the C stack. This involves resuming the last incomplete suspended computation with all references to the completed suspension replaced by the computed value. remember hd(S); recover S, E, C and D from hd(D); throughout S and E replace all occurrences of the suspension that is hd(tl(S)) by what was hd(S)

19. CS321 Functional Programming 2 © JAS 2005 6-19 S E C D (y,6) (λx.4) y (y,6) ap, (λx.4), y 6 (y,6) ap, (λx.4) 6,cl(4,x,((y,6)) (y,6) ap (y,6),(x,6) 4 ((),(y,6),(),()) 4 (y,6),(x,6) ((),(y,6),(),()) 4 (y,6) Consider an example evaluation using the standard eager machine

20. CS321 Functional Programming 2 © JAS 2005 6-20 S E C D (y,6) (λx.4) y su(y,(y,6)) (y,6) ap, (λx.4) su(y,(y,6)), cl(4,x,((y,6)) (y,6) ap (y,6),(x,su(y,(y,6))) 4 ((),(y,6),(),()) 4 (y,6),(x,su(y,(y,6))) ((),(y,6),(),()) 4 (y,6) The same example evaluation using a lazy machine