Re: [gambit-list] Proper usage of ffi structs

Aha.

Very well, then I guess it's good time to ask Marc how this stuff works really. :))

Marc, would you feel like enlightening us on this one - how use and reclaim of non-pointered FFI structs works out?

Thanks, Mikael

2013/8/22 Chris Mueller ruunsmail@gmail.com

Good point.

But gsc / gcc are reclaiming ___result_voidstar is of type void* and has no elements x and y therefore.

Manually casting to struct point* is of course possible, but will result in an expected segmentation fault during the execution of the script.

On 22.08.2013 08:34, Mikael wrote:

...

What about you presume that ___result_voidstar is already allocated, and you just access/mutate it? I.e.

(define make-point (c-lambda (int int) struct-point "___result_voidstar->.x = ___arg1; ___result_voidstar->y = ___arg2;"))

?

2013/8/22 Chris Mueller <ruunsmail@gmail.com <mailto:ruunsmail@gmail.com

...

I'm opening a new subject for the ffi struct issue to sum up the

current

...

discussion:

In 19.1 "The Mapping of types between C and Scheme" of the gambit
documentation is mentioned:

"In the case of the struct, union and type types, the default

function

...

reclaims the copy on the C heap referenced by the internal pointer

(when

...

using a C++ compiler this is done using ‘delete
(type*)internal-pointer’, which calls the destructor of type if it

is a

...

class) and returns ‘___FIX(___NO_ERR)’. In many situations the

default

...

release-function will perform the appropriate cleanup for the foreign
type."

If i interpret this correctly, a declared struct type in a ffi

provides

...

a default release function that automatically calls a destructor

(used

...

with g++) and frees the corresponding memory.

Unfortanetly, i'm little lost to use it correctly and experiment
currently heavily to find the right way.


Assume a type:

(c-declare "struct point { int x; int y; };")
(c-define-type struct-point (struct "point"))


Assume a test programm that allocates a point 1_000_000_000 times.

(define (alloc-point times)
    (let loop ((i 0))
       (if (< i times)
         (let ((a (make-point 0 0)))
           (loop (+ i 1)))))))

(alloc-point 1000000000)


How do we implement (make-point x y) to use a ffi struct type with a
default release function in Gambit?

1) With a manually allocated pointer it will leak memory.

(define make-point
    (c-lambda (int int) struct-point
      "point* a = new Point;
       a->x = ___arg1;
       a->y = ___arg2;
       ___result_voidstar = a;"))

2) With a stack-allocated value it will segfault immediately.

(define make-point
    (c-lambda (int int) struct-point
      "point a;
       a.x = ___arg1;
       a.y = ___arg2;
       ___result = a;"))

3) If we give the address of a stack-allocated value to
___result_voidstar the allocation loop is running with constant

memory

...

space but the programm will segfault after quiting with CTRL + X.

(define make-point
    (c-lambda (int int) struct-point
      "point a;
       a.x = ___arg1;
       a.y = ___arg2;
       ___result_voidstar = &a;"))


For more details, see discussion:

https://mercure.iro.umontreal.ca/pipermail/gambit-list/2013-August/006950.ht...

...

Chris
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