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Re: Pure (side-effect-free) calls into c/c++?
From: |
Taylan Kammer |
Subject: |
Re: Pure (side-effect-free) calls into c/c++? |
Date: |
Sun, 12 Jan 2020 04:21:49 +0100 |
User-agent: |
Mozilla/5.0 (Windows NT 10.0; WOW64; rv:68.0) Gecko/20100101 Thunderbird/68.3.1 |
On 12.01.2020 03:12, Linas Vepstas wrote:
> To answer my own question, it appears doable with a very simply macro, then:
>
> (define (bar x)
> (format #t "Called bar with ~A\n" x)
> (+ x 1))
>
> (define (memoize FUNC)
> "
> memoize a function FUNC which takes a single int as argument
> "
> (define cache (make-hash-table))
> (define (int-hash INT SZ) (modulo INT SZ))
> (define (int-assoc INT ILIST)
> (find (lambda (pr) (equal? INT (car pr))) ILIST))
>
> (lambda (ITEM)
> (define val (hashx-ref int-hash int-assoc cache ITEM))
> (if val val
> (let ((fv (FUNC ITEM)))
> (hashx-set! int-hash int-assoc cache ITEM fv)
> fv)))
> )
>
> (define bar-memo (memoize bar))
>
> (define-syntax foo
> (syntax-rules ()
> ((foo exp)
> (if (symbol? (quote exp))
> (begin (display "its a symb\n") (bar exp))
> (begin (display "no its not\n") (bar-memo exp))))))
Using (quote exp) in the macro output to determine if the input was a
symbol is... smart I guess! There's a problem in your approach though:
the memoization and the lookup of the memoized value will happen at
run-time, because your macro is merely emitting code that calls the
function with the memoization cache.
This also means, of course, that checking whether the argument was a
symbol is useless.
In your (lambda (ITEM) ...) definition, insert some (display ...) lines
to print "was already memoized" or "will now be memoized" and you will
see what I mean. Here's how calls to the function might then look:
scheme> (define x 66)
scheme> (bar-memo x)
will now be memoized
$1 = 67
scheme> (bar-memo x)
was already memoized
$2 = 67
scheme> (set! x 42)
scheme> (bar-memo x)
will now be memoized
$3 = 43
scheme> (bar-memo x)
was already memoized
$4 = 43
It's important to understand that Scheme uses "pass-by-value" semantics
in procedure calls. That means that a procedure doesn't know how the
values it receives were computed -- whether they were constants, or
variable references, or the result of another function call.
Consider:
(foo 42)
(let ((x 42))
(foo x))
(foo (+ 21 21))
In all three cases, foo receives the same value, 42, and doesn't know
how it arrived.
I'm not sure if memoization is really what you want, since it's a
run-time caching mechanism.
It might be possible to create a sort of "compile-time memoization" but
it would be quite complicated. For each constant value the macro
receives, it would emit a global variable definition that will be bound
to the call to the actual function at run-time, and then each call with
the same constant would emit a reference to that variable.
So the following:
(display (f-memo 42))
(display (f-memo 66))
(display (f-memo 42))
(display (f-memo 66))
would magically emit code like:
(define _x1 (f 42))
(define _x2 (f 66))
(display _x1)
(display _x2)
(display _x1)
(display _x2)
But I'm not sure how I'd write that hypothetical f-memo macro. You
can't really make a macro-call deep within some code emit a top-level
variable definition. All I can imagine are some pretty dirty methods
that probably aren't worth the complexity.
- Taylan
Re: Pure (side-effect-free) calls into c/c++?, Linus Björnstam, 2020/01/11