I'm being greedy here ... how can I modify this code so that if I have:
(c-define-type Foo:Bar "FooBar")
(define-array-allocator-and-accessors Foo:Bar)
will result in using "FooBar" in the generated *.c file rather than Foo:Bar
Here is another, more generic, approach which uses a macro to generate the various functions:
On 15-Feb-09, at 12:29 AM, lowly coder wrote:
... wow ... thanks!
On Sat, Feb 14, 2009 at 9:14 PM, Marc Feeley <feeley@iro.umontreal.ca> wrote:
On 14-Feb-09, at 11:58 PM, lowly coder wrote:
question 1:
Is it possible to write gambit code to produce the equiv of:
int foo[200]; ?
the current best I have in mind is:
(let ((foo (malloc (* 200 8)))) ... )
end-of-c-declare
(c-declare #<<end-of-c-declare
#include <stdlib.h>
)
(define-macro (define-array-allocator-and-accessors type)
(define (sym . syms)
(string->symbol (apply string-append
(map symbol->string syms))))
(let ((type-str (symbol->string type)))
`(begin
(define ,(sym 'alloc- type '-array)
(c-lambda (int) (pointer ,type)
,(string-append
"___result_voidstar = malloc( ___arg1 * sizeof("
type-str
") );")))
(define ,(sym 'free- type '-array)
(c-lambda ((pointer ,type)) void
"free( ___arg1 );"))
(define ,(sym type '-array-ref)
(c-lambda ((pointer ,type) int) ,type
"___result = ___arg1[___arg2];"))
(define ,(sym type '-array-set!)
(c-lambda ((pointer ,type) int ,type) void(define-array-allocator-and-accessors int)
"___arg1[___arg2] = ___arg3;")))))
;; test it:
(define-array-allocator-and-accessors double)
(define a (alloc-int-array 5))
(define b (alloc-double-array 10))(double-array-set! b 1 1.5)
(int-array-set! a 1 111)
(int-array-set! a 2 222)
(pp (+ (int-array-ref a 1) (int-array-ref a 2))) ;; 333
(double-array-set! b 2 2.0)
(pp (+ (double-array-ref b 1) (double-array-ref b 2))) ;; 3.5
(free-int-array a)
(free-double-array b)