OK, I see. The expression is an error without the assignment to a, but
the assignment creates a race condition that can result in a self-reference.
I guess it is possible to detect the situation when all the non-idle
threads are waiting for a pyx, and halt everything then. Like with
jbreak, what state would the threads be in?
Henry Rich
On 4/13/2022 5:41 PM, Elijah Stone wrote:
I am pretty sure it is a deadlock. Here's an alternate expression
which requires no delay:
a=: {{> b=: >t.''a}}t.'' 0
>a
[hang]
As I see it, a cannot make progress until b finishes (because a is
waiting for b), but b cannot make progress until a finishes (because b
is waiting for a), and since it is necessary for a running thread to
make progress before it can finish, neither a nor b can make progress.
On Wed, 13 Apr 2022, Henry Rich wrote:
I don't think this is a deadlock. It's just an infinite recursion,
made painfully slow by the delay.
Every task is started with enough resources to finish the task. As
long as the OS has some way to ensure that no task will starve,
deadlock should be impossible.
That was my reasoning when I designed this. I sure hope it's right.
Henry Rich
On 4/13/2022 1:12 AM, Elijah Stone wrote:
a=: {{> ". 'b' [ 6!:3]2}} t.'' 0
b=: {{> ". 'a' [ 6!:3]2}} t.'' 0
>a
[hang]
actually, it's enough to make it wait on itself:
a=: {{> ". 'a' [ 6!:3]2}}t.'' 0
>a
[hang]
Probably it should be an error. E.G. pthread_join will return
EDEADLK when it encounters an analogous situation.
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