Nathaniel Smith added the comment:
On further thought, I think the way I'd write a test for this is:
(1) add a testing primitive that waits for N instructions and then injects a
SIGINT. Probably this would require tweaking the definition of
Py_MakePendingCalls like I described in my previous comment, and then some code
like:
int sigint_timebomb(void* count_ptr) {
intptr_t count = (intptr_t) count_ptr;
if (count == 0)
raise(SIGINT);
else
Py_AddPendingCall(sigint_timebomb, (void*) (count - 1));
return 0;
}
(2) write a test like:
reached_crashpad = False
lock = threading.Lock()
i = 0
while not reached_crashpad:
i += 1
try:
sigint_timebomb(i)
with lock:
1 + 1
# this loop keeps executing instructions until any still-armed
# timebombs go off
while True:
reached_crashpad = True
except KeyboardInterrupt:
# key invariant: signal can't leave the lock held
assert not lock.locked()
else:
assert False # can't happen
The idea here is that this sets off SIGINTs at every possible place throughout
the execution of the 'with lock', while checking that the lock is always
released, and without needing to hard code any information at all about what
bytecode the compiler generated.
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue29988>
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