Optimizer

[email protected] wrote:

>>(define (f a b) (+ a (* b 2)))
>>    
>>
>>(procedure-body f) => (lambda (a b) (+ a (* b 2)))
>>    
>>
>Your best bet is to modify the optimizer, or create your own, that emits 
>the following code:
>
>(define lhs (lambda formals body)) =>
>(define lhs (let ([tmp (lambda formals xbody)])
>              (procedure-property tmp 'body 'xbody)
>              tmp))
>
After talking with Matthias on #sisc I understand this much better but I 
still have a long way to go, I think.

1. You suggest that I write a case for the match in |opt| that handles 
"(define lhs (lambda formals body))"?
Yes, and it should probably call opt:lambda to optimize the lambda thats 
there so you don't loose out on optimizations.

2. Where did xbody come from? 
Typo because I was writing a rough impl.  More like:

  ((define ,lhs (lambda ,formals ,body))
   (let ([new-lambda (opt:lambda formals body state)])
     `(define ,lhs ,((lambda (tmp)
	              (begin
                        (set-procedure-property! tmp 'body ,body)
                        tmp))
                     ,new-lambda))))

3. Did you mean that handler should use |set-procedure-property| ?  
Otherwise, how is the property set?
Right, my bad.
4. What's the match expression? "((define ,lhs ,[lambda formals body])" ?

In the match cases, ,var means match anything an assign it to the 
variable "var".  ,[var] means recurse on the expression (ie if your in 
opt, call (opt var), and bind the results to "var".  ,[var1 var2] is the 
same, but expects multiple return values, assigning the first to var1 
and the second to var2.

5. Do I need a separate case for "(define (foo bar) (shoo bar))" ?  I 
need to have the 'body property on those too.

No, the syntax expander has already converted everything to simple 
expressions.

6. Do I need to modify both |opt| and |scan|?
No.  Scan is responsible for collecting information about lexical 
variables, variables that are set!'ed, etc.  This information is 
passed to opt does the actual transformations.

	Scott

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