(async/go (async/chan)) gc ?

2014-01-30 Thread t x
Hi,

  Consider this block of code:

  (async/go (async/chan))

  This creates a go-thread and a go-chan.

  The go-thread is linked from the list of blocked go-threads.

  The go-chan is linked from the go-thread.

  Thus, did we just create a go-thread and a go-chan which is NOT gc-ed at all?

Thanks!

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Re: (async/go (async/chan)) gc ?

2014-01-30 Thread Timothy Baldridge
The key thing to remember in all questions of core.async GC is that there
is no global blocked gos queue. Instead, the blocked gos are attached
directly to the channel. When the channel is GC'd all attached gos are GC'd
as well.

So in this case, it's actually even simpler.the go never blocks. What
this code is doing is creating a channel to return from the go. This
channel has a buffer size of 1. It then puts a channel into the channel.
The go then exits and is GC'd. Once the return channel is discarded by the
caller of this code, both channels are also GC'd

Timothy


On Thu, Jan 30, 2014 at 11:24 AM, t x txrev...@gmail.com wrote:

 Hi,

   Consider this block of code:

   (async/go (async/chan))

   This creates a go-thread and a go-chan.

   The go-thread is linked from the list of blocked go-threads.

   The go-chan is linked from the go-thread.

   Thus, did we just create a go-thread and a go-chan which is NOT gc-ed at
 all?

 Thanks!

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Re: (async/go (async/chan)) gc ?

2014-01-30 Thread t x
 The key thing to remember in all questions of core.async GC is that there is
 no global blocked gos queue. Instead, the blocked gos are attached
 directly to the channel. When the channel is GC'd all attached gos are GC'd
 as well.

This is the source of all my mis-understanding. Thanks for explaining!



 So in this case, it's actually even simpler.the go never blocks. What
 this code is doing is creating a channel to return from the go. This channel
 has a buffer size of 1. It then puts a channel into the channel. The go then
 exits and is GC'd. Once the return channel is discarded by the caller of
 this code, both channels are also GC'd

 Timothy


 On Thu, Jan 30, 2014 at 11:24 AM, t x txrev...@gmail.com wrote:

 Hi,

   Consider this block of code:

   (async/go (async/chan))

   This creates a go-thread and a go-chan.

   The go-thread is linked from the list of blocked go-threads.

   The go-chan is linked from the go-thread.

   Thus, did we just create a go-thread and a go-chan which is NOT gc-ed at
 all?

 Thanks!

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Re: (async/go (async/chan)) gc ?

2014-01-30 Thread t x
Following up on this. Is the following correct:

* async/thread creates Java threads -- these _are_ root gc objects

* async/go creates go blocks, which in reality, is just a state
machine. go-blocks are NOT root gc objects

* go-blocks do _NOT_ have stackframes -- they're just a simple, flat
state machine created via inversion of control at !, !, and alt!
locations

* when a go-block blocks, it gets appended to either the put or the
take queue of the channel it's blocking on

Thus, in the following code:

(let [c (async/chan)
  d (async/go (! c))] )

what happens is the following thing:

* we create a channel c
* we create a go-block, who returns to a channel d
* the go block hits (! c), then gets ioc-ed into a state machine
* this state machine is put on the take-list of c

Then, the GC collects c and d -- and the go-block vanishes, since
it's just a state machine attached to c


Can you please tell me if

(1) all above statements are correct or
(2) correct the wrong ones in place?

Thanks!


On Thu, Jan 30, 2014 at 11:06 AM, t x txrev...@gmail.com wrote:
 The key thing to remember in all questions of core.async GC is that there is
 no global blocked gos queue. Instead, the blocked gos are attached
 directly to the channel. When the channel is GC'd all attached gos are GC'd
 as well.

 This is the source of all my mis-understanding. Thanks for explaining!



 So in this case, it's actually even simpler.the go never blocks. What
 this code is doing is creating a channel to return from the go. This channel
 has a buffer size of 1. It then puts a channel into the channel. The go then
 exits and is GC'd. Once the return channel is discarded by the caller of
 this code, both channels are also GC'd

 Timothy


 On Thu, Jan 30, 2014 at 11:24 AM, t x txrev...@gmail.com wrote:

 Hi,

   Consider this block of code:

   (async/go (async/chan))

   This creates a go-thread and a go-chan.

   The go-thread is linked from the list of blocked go-threads.

   The go-chan is linked from the go-thread.

   Thus, did we just create a go-thread and a go-chan which is NOT gc-ed at
 all?

 Thanks!

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 zero-they had no way to indicate successful termination of their C
 programs.
 (Robert Firth)

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Re: (async/go (async/chan)) gc ?

2014-01-30 Thread Jozef Wagner
Seems OK. Note that code inside go blocks is transformed into state machine
'object' immediatelly, not waiting for hitting !, as go is a macro.


On Thu, Jan 30, 2014 at 8:36 PM, t x txrev...@gmail.com wrote:

 Following up on this. Is the following correct:

 * async/thread creates Java threads -- these _are_ root gc objects

 * async/go creates go blocks, which in reality, is just a state
 machine. go-blocks are NOT root gc objects

 * go-blocks do _NOT_ have stackframes -- they're just a simple, flat
 state machine created via inversion of control at !, !, and alt!
 locations

 * when a go-block blocks, it gets appended to either the put or the
 take queue of the channel it's blocking on

 Thus, in the following code:

 (let [c (async/chan)
   d (async/go (! c))] )

 what happens is the following thing:

 * we create a channel c
 * we create a go-block, who returns to a channel d
 * the go block hits (! c), then gets ioc-ed into a state machine
 * this state machine is put on the take-list of c

 Then, the GC collects c and d -- and the go-block vanishes, since
 it's just a state machine attached to c


 Can you please tell me if

 (1) all above statements are correct or
 (2) correct the wrong ones in place?

 Thanks!


 On Thu, Jan 30, 2014 at 11:06 AM, t x txrev...@gmail.com wrote:
  The key thing to remember in all questions of core.async GC is that
 there is
  no global blocked gos queue. Instead, the blocked gos are attached
  directly to the channel. When the channel is GC'd all attached gos are
 GC'd
  as well.
 
  This is the source of all my mis-understanding. Thanks for explaining!
 
 
 
  So in this case, it's actually even simpler.the go never blocks.
 What
  this code is doing is creating a channel to return from the go. This
 channel
  has a buffer size of 1. It then puts a channel into the channel. The go
 then
  exits and is GC'd. Once the return channel is discarded by the caller of
  this code, both channels are also GC'd
 
  Timothy
 
 
  On Thu, Jan 30, 2014 at 11:24 AM, t x txrev...@gmail.com wrote:
 
  Hi,
 
Consider this block of code:
 
(async/go (async/chan))
 
This creates a go-thread and a go-chan.
 
The go-thread is linked from the list of blocked go-threads.
 
The go-chan is linked from the go-thread.
 
Thus, did we just create a go-thread and a go-chan which is NOT
 gc-ed at
  all?
 
  Thanks!
 
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To 

Re: (async/go (async/chan)) gc ?

2014-01-30 Thread t x
Great! Thanks for verifying.

On Thu, Jan 30, 2014 at 11:46 AM, Jozef Wagner jozef.wag...@gmail.com wrote:
 Seems OK. Note that code inside go blocks is transformed into state machine
 'object' immediatelly, not waiting for hitting !, as go is a macro.


 On Thu, Jan 30, 2014 at 8:36 PM, t x txrev...@gmail.com wrote:

 Following up on this. Is the following correct:

 * async/thread creates Java threads -- these _are_ root gc objects

 * async/go creates go blocks, which in reality, is just a state
 machine. go-blocks are NOT root gc objects

 * go-blocks do _NOT_ have stackframes -- they're just a simple, flat
 state machine created via inversion of control at !, !, and alt!
 locations

 * when a go-block blocks, it gets appended to either the put or the
 take queue of the channel it's blocking on

 Thus, in the following code:

 (let [c (async/chan)
   d (async/go (! c))] )

 what happens is the following thing:

 * we create a channel c
 * we create a go-block, who returns to a channel d
 * the go block hits (! c), then gets ioc-ed into a state machine
 * this state machine is put on the take-list of c

 Then, the GC collects c and d -- and the go-block vanishes, since
 it's just a state machine attached to c


 Can you please tell me if

 (1) all above statements are correct or
 (2) correct the wrong ones in place?

 Thanks!


 On Thu, Jan 30, 2014 at 11:06 AM, t x txrev...@gmail.com wrote:
  The key thing to remember in all questions of core.async GC is that
  there is
  no global blocked gos queue. Instead, the blocked gos are attached
  directly to the channel. When the channel is GC'd all attached gos are
  GC'd
  as well.
 
  This is the source of all my mis-understanding. Thanks for explaining!
 
 
 
  So in this case, it's actually even simpler.the go never blocks.
  What
  this code is doing is creating a channel to return from the go. This
  channel
  has a buffer size of 1. It then puts a channel into the channel. The go
  then
  exits and is GC'd. Once the return channel is discarded by the caller
  of
  this code, both channels are also GC'd
 
  Timothy
 
 
  On Thu, Jan 30, 2014 at 11:24 AM, t x txrev...@gmail.com wrote:
 
  Hi,
 
Consider this block of code:
 
(async/go (async/chan))
 
This creates a go-thread and a go-chan.
 
The go-thread is linked from the list of blocked go-threads.
 
The go-chan is linked from the go-thread.
 
Thus, did we just create a go-thread and a go-chan which is NOT
  gc-ed at
  all?
 
  Thanks!
 
  --
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  that-lacking
  zero-they had no way to indicate successful termination of their C
  programs.
  (Robert Firth)
 
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