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I'm somewhat confused regarding dynamical scoping, specifically what happens when a formal parameter and a free variable share a name.

For example

(define x 1)
(define f (lambda (x) x) )
(f 2)

If this code is compiled and evaluated using dynamic scoping, what would be the input? 2 or 3?

While it seems as if the assignment (x = 2) from the parameter is the most 'recent' so it should be '2', some have told me the answer is 1 (and others said its 2. Everyone is confused).

{I know scheme and most languages use lexical scoping, but tell that to my professor}

I would appreciate any help I can get.

dippas
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Grim Oir-lee
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1 Answers1

1

There is no dynamic scoping. It is either "lexical scoping" or "dynamic extent". But, putting that aside, it's 2:

(f 2) 
= 
( (lambda (x) x) 2 ) 
= 
(begin (set! old-x x)           ; save old value of formal param
       (set! x 2)               ; set param to argument's value
       (set! return-value       ; evaluate the body
             x)                 ;   and save the result
       (set! x old-x)           ; restore the param's old value
       return-value )           ; return the result

The translation must use unique names for the temporary variables throughout, to prevent any mix-ups. Here it's old-x and return-value but it could be any name not used anywhere else in the translated program code.

Will Ness
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