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fix `sync/[timeout/]enable-break' semaphore shortcut

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The shortcut could be triggered in a bad case (first
argument as `#f' in non-timeout mode) and returned the
wrong result (void instead of the semaphore).
This commit is contained in:
Matthew Flatt 2013-03-24 10:49:49 -06:00
parent b041a151e6
commit c3266ef685
2 changed files with 18 additions and 3 deletions
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6
collects/tests/racket/sync.rktl
View File

@ -1236,4 +1236,10 @@
;; ----------------------------------------
(err/rt-test (sync/enable-break #f (make-semaphore 1)))
(test #t semaphore? (sync/enable-break (make-semaphore 1)))
(test #t semaphore? (sync/timeout/enable-break #f (make-semaphore 1)))
;; ----------------------------------------
(report-errs)

15
src/racket/src/thread.c
View File

@ -6759,9 +6759,18 @@ Scheme_Object *scheme_sync_timeout(int argc, Scheme_Object *argv[])
static Scheme_Object *do_scheme_sync_enable_break(const char *who, int with_timeout, int tailok, int argc, Scheme_Object *argv[])
{
if (argc == 2 && SCHEME_FALSEP(argv[0]) && SCHEME_SEMAP(argv[1])) {
scheme_wait_sema(argv[1], -1);
return scheme_void;
Scheme_Object *sema;
if (with_timeout && (argc == 2) && SCHEME_FALSEP(argv[0]) && SCHEME_SEMAP(argv[1]))
sema = argv[1];
else if (!with_timeout && (argc == 1) && SCHEME_SEMAP(argv[0]))
sema = argv[0];
else
sema = NULL;
if (sema) {
scheme_wait_sema(sema, -1);
return sema;
}
return do_sync(who, argc, argv, 1, with_timeout, tailok);