4 ms·

I think that's just normal code, nothing that affects the runtime, that Promise is just a pony actor and abides to the same rules, with the same causal messagin

by ndr 5y ago

I think that's just normal code, nothing that affects the runtime, that Promise is just a pony actor and abides to the same rules, with the same causal messaging guarantees.

gpderetta 5y ago
Let's say I'm writing a trivial proxy which simply forwards data unchanged between two sockets. This is my pseudocode: while true: x = read s1 write s2, x y = read s2 write s1, y This can deadlock. If for example s1 stops sending data, read(s1) blocks, so the proxy won't forward any data from s2 back to s1, potentially leading to a full pipeline deadlock. Now, the above can be expressed in purely non blocking fashion as follow : letrec f = async_read s1, \x -> async_write s2, x \-> async_read s2, \y -> async_write s1, y, f It still deadlocks exactly as the previous program. I'm not going to learn pony for the sake of an example, but I'm 97.5% sure you can express the above in Pony, except you do not pass the continuation directly to read and writes but to the returned promise. Sure, the pony runtime will still work and technically the actor could potentially handle other messages, but that's not very useful for the end user that has to deal with a deadlocked pipeline. So claiming that pony is deadlock-free might be true in a trivial sense, but in practice it is at the very least misleading and potentially dangerous. edit: it is of course possible to write the program in such a way that it doesn't deadlock, the point is that if pony was truly deadlock free it would reject all programs that can deadlock. edit2: and of course, if one were to show that the above program is in fact not expressible in pony, it would be a great result and I would want to know about!
- ndr 5y ago
  I think you might not be able to write that program, there simply is no way to block in Pony (modulo FFI). The scheduler wakes up actors that have messages (in their local/private queue) to process, all actors can do is to consume their queue and send (wait-free) messages to other actors if they want to. From your example "read X" would be reading from the actor's queue, so it would not be blocking. If the message queue is empty that actor won't be scheduled, no blocking there. That "while true" would not be writeable as such, you'd have to have two "behaviors" (in Pony terms, you can imagine it's an event handler) one that process the writing of s1 and one for the writing of s2, and "on read" they'd do the write. There'd be no "while true". > I'm not going to learn pony for the sake of an example ... > edit2: and of course, if one were to show that the above program is in fact not expressible in pony, it would be a great result and I would want to know about! I might have misunderstood what your example was supposed to do, but I suspect you'd find it interesting to try it out.
  - gpderetta 5y ago
    The async_{read,write} variant never blocks nor has a while loop, it is just a four continuations recursively tail calling each other. Yet it deadlocks (as it is a purely mechanical transformation of the synchronous variant). It still might be possible that the precise pony scoping rules for closures prevent writing that variant, but I would be surprised.

by ndr 5y ago

I think that's just normal code, nothing that affects the runtime, that Promise is just a pony actor and abides to the same rules, with the same causal messaging guarantees.

gpderetta 5y ago

Let's say I'm writing a trivial proxy which simply forwards data unchanged between two sockets. This is my pseudocode: while true: x = read s1 write s2, x y = read s2 write s1, y This can deadlock. If for example s1 stops sending data, read(s1) blocks, so the proxy won't forward any data from s2 back to s1, potentially leading to a full pipeline deadlock. Now, the above can be expressed in purely non blocking fashion as follow : letrec f = async_read s1, \x -> async_write s2, x \-> async_read s2, \y -> async_write s1, y, f It still deadlocks exactly as the previous program. I'm not going to learn pony for the sake of an example, but I'm 97.5% sure you can express the above in Pony, except you do not pass the continuation directly to read and writes but to the returned promise. Sure, the pony runtime will still work and technically the actor could potentially handle other messages, but that's not very useful for the end user that has to deal with a deadlocked pipeline. So claiming that pony is deadlock-free might be true in a trivial sense, but in practice it is at the very least misleading and potentially dangerous. edit: it is of course possible to write the program in such a way that it doesn't deadlock, the point is that if pony was truly deadlock free it would reject all programs that can deadlock. edit2: and of course, if one were to show that the above program is in fact not expressible in pony, it would be a great result and I would want to know about!

ndr 5y ago
I think you might not be able to write that program, there simply is no way to block in Pony (modulo FFI). The scheduler wakes up actors that have messages (in their local/private queue) to process, all actors can do is to consume their queue and send (wait-free) messages to other actors if they want to. From your example "read X" would be reading from the actor's queue, so it would not be blocking. If the message queue is empty that actor won't be scheduled, no blocking there. That "while true" would not be writeable as such, you'd have to have two "behaviors" (in Pony terms, you can imagine it's an event handler) one that process the writing of s1 and one for the writing of s2, and "on read" they'd do the write. There'd be no "while true". > I'm not going to learn pony for the sake of an example ... > edit2: and of course, if one were to show that the above program is in fact not expressible in pony, it would be a great result and I would want to know about! I might have misunderstood what your example was supposed to do, but I suspect you'd find it interesting to try it out.
- gpderetta 5y ago
  The async_{read,write} variant never blocks nor has a while loop, it is just a four continuations recursively tail calling each other. Yet it deadlocks (as it is a purely mechanical transformation of the synchronous variant). It still might be possible that the precise pony scoping rules for closures prevent writing that variant, but I would be surprised.