digitalmars.D.learn - Threads not independent...
- Larry Cowan (39/39) Apr 04 2005 | module test;
- Larry Cowan (2/41) Apr 04 2005
- Regan Heath (29/70) Apr 04 2005 In the above all the unnamed delegates share the same stack. Containing ...
- Larry Cowan (14/87) Apr 04 2005 Its not "free"ed, but it is reused, maybe j then uses the same place - t...
- Regan Heath (23/130) Apr 04 2005 Good point. This seems likely to me.
- Larry Cowan (75/130) Apr 04 2005 +| int test = 99;
- Regan Heath (39/39) Apr 04 2005 Try this:
- Regan Heath (40/40) Apr 04 2005 Alternate solution using seperate object, not derived from Thread:
- Regan Heath (24/24) Apr 04 2005 Ooops. One important correction is that the 'allDone' variable actually ...
- Larry Cowan (2/41) Apr 04 2005
- Ben Hinkle (14/20) Apr 04 2005 Glad to see someone using the Locks lib. When I was testing things out I...
- Ben Hinkle (13/39) Apr 04 2005 I should add that if you are on Linux and you start seeing seg-v's on
- Carlos Santander B. (8/55) Apr 04 2005 The problem is, you're asking the threads to print the value of "i", but...
| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[i].start(); | | allDone.wait(); | printf("All done.\n"); |} prints: |Starting threads. | thread 5 | thread 5 | thread 5 | thread 5 | thread 5 |All done. How do I best do what I obviously want to? I can thnk of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack? Do I have to make one for them on the heap to actually get independent activity, then have them pick up an index from a locked counter?
Apr 04 2005
Corrected error in code given, bad typing... In article <d2sdsg$9ca$1 digitaldaemon.com>, Larry Cowan says...| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[j].start(); | | allDone.wait(); | printf("All done.\n"); |} prints: |Starting threads. | thread 5 | thread 5 | thread 5 | thread 5 | thread 5 |All done. How do I best do what I obviously want to? I can think of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack? Do I have to make one for them on the heap to actually get independent activity, then have them pick up an index from a locked counter?
Apr 04 2005
On Mon, 4 Apr 2005 22:08:16 +0000 (UTC), Larry Cowan <Larry_member pathlink.com> wrote:| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[i].start(); | | allDone.wait(); | printf("All done.\n"); |} prints: |Starting threads. | thread 5 | thread 5 | thread 5 | thread 5 | thread 5 |All done. How do I best do what I obviously want to? I can thnk of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack?In the above all the unnamed delegates share the same stack. Containing the same 'i' variable. The dodgy bit, from where I'm sitting is that 'i' only exists for the duration of the for loop, so by the time you 'start' the threads... I must be missing something here? Perhaps the stack is not 'free'd (?) unless no more 'refences'(?) to it exist?Do I have to make one for them on the heap to actually get independent activity, then have them pick up an index from a locked counter?I would write a class i.e. class Foo { int index; this(int i) { index = i; } .. run() .. } then | for (int i=0; i < 5 ;i++) | t[i] = new Thread(new Foo(i).run); (not sure if the above is legal, but you get the idea?) This should create a delegate that's object pointer points to a Foo class which has it's own 'index' integer, not shared between all the threads. You could instead derive a clas from thread eg. class Foo : Thread { int index; this(int i, ?delegate?) { super(?delegate?); index = i; } ~this() { allDone.countDown(); } // or maybe put this in the 'stop'(?) type method? } "?delegate?" above as I cannot remember the syntax off hand. Regan
Apr 04 2005
See below. In article <opsoqa0osi23k2f5 nrage.netwin.co.nz>, Regan Heath says...On Mon, 4 Apr 2005 22:08:16 +0000 (UTC), Larry Cowan <Larry_member pathlink.com> wrote:Its not "free"ed, but it is reused, maybe j then uses the same place - the effect would be the same, but I can test this. Anyway, I know what is happening re the main() sction of the stack, but I was asking for an idea about how best to staticize the index for each thread, and I will try your "new"ing it idea, but the ref is still in main()'s stack area and may have the same problem.| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[i].start(); | | allDone.wait(); | printf("All done.\n"); |} prints: |Starting threads. | thread 5 | thread 5 | thread 5 | thread 5 | thread 5 |All done. How do I best do what I obviously want to? I can thnk of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack?In the above all the unnamed delegates share the same stack. Containing the same 'i' variable. The dodgy bit, from where I'm sitting is that 'i' only exists for the duration of the for loop, so by the time you 'start' the threads... I must be missing something here? Perhaps the stack is not 'free'd (?) unless no more 'refences'(?) to it exist?Also I wanted assurance that when the threads actually start they have independent stacks without my help - not just as if they were called by main() and got the next stack frame - each getting the same one.Do I have to make one for them on the heap to actually get independent activity, then have them pick up an index from a locked counter?I would write a class i.e. class Foo { int index; this(int i) { index = i; } .. run() .. } then | for (int i=0; i < 5 ;i++) | t[i] = new Thread(new Foo(i).run); (not sure if the above is legal, but you get the idea?) This should create a delegate that's object pointer points to a Foo class which has it's own 'index' integer, not shared between all the threads. You could instead derive a clas from thread eg. class Foo : Thread { int index; this(int i, ?delegate?) { super(?delegate?); index = i; } ~this() { allDone.countDown(); } // or maybe put this in the 'stop'(?) type method? } "?delegate?" above as I cannot remember the syntax off hand. ReganI'll report back after some more experimenting... -Larry
Apr 04 2005
On Mon, 4 Apr 2005 22:42:21 +0000 (UTC), Larry Cowan <Larry_member pathlink.com> wrote:In article <opsoqa0osi23k2f5 nrage.netwin.co.nz>, Regan Heath says...Good point. This seems likely to me.On Mon, 4 Apr 2005 22:08:16 +0000 (UTC), Larry Cowan <Larry_member pathlink.com> wrote:Its not "free"ed, but it is reused, maybe j then uses the same place - the effect would be the same, but I can test this.| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[i].start(); | | allDone.wait(); | printf("All done.\n"); |} prints: |Starting threads. | thread 5 | thread 5 | thread 5 | thread 5 | thread 5 |All done. How do I best do what I obviously want to? I can thnk of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack?In the above all the unnamed delegates share the same stack. Containing the same 'i' variable. The dodgy bit, from where I'm sitting is that 'i' only exists for the duration of the for loop, so by the time you 'start' the threads... I must be missing something here? Perhaps the stack is not 'free'd (?) unless no more 'refences'(?) to it exist?Anyway, I know what is happening re the main() sction of the stack, but I was asking for an idea about how best to staticize the index for each thread, and I will try your "new"ing it idea, but the ref is still in main()'s stack area and may have the same problem.No, my ideas do not use main()'s stack area. Each class is allocated on the heap, and the reference to the class is not stored except in the delegate passed to the threads constructor. (or in t[] if you extend Thread with your own class - which still means it is not stored in mains stack area - only the t[] reference is). Either way the stack frame of the thread code is not main() it is the stack frame of the function in the class.IIRC creating an unnamed delegate assigns the current stack frame to the delegates object pointer. So, creating an unnamed delegate in "main()" assigns mains stack frame to the delegate. A class method delegate has it's object pointer assigned to the class instance. The stack frame for a class delegate is the stack frame created when the method is called.Also I wanted assurance that when the threads actually start they have independent stacks without my help - not just as if they were called by main() and got the next stack frame - each getting the same one.Do I have to make one for them on the heap to actually get independent activity, then have them pick up an index from a locked counter?I am interested to know what you find. I have done very little experimentation with threads myself, so my advice may not be entirely acurate. IIRC Russ Lewis is the person I see most often posting about delegates and stack frames, his advice you can trust (I believe). ReganI would write a class i.e. class Foo { int index; this(int i) { index = i; } .. run() .. } then | for (int i=0; i < 5 ;i++) | t[i] = new Thread(new Foo(i).run); (not sure if the above is legal, but you get the idea?) This should create a delegate that's object pointer points to a Foo class which has it's own 'index' integer, not shared between all the threads. You could instead derive a clas from thread eg. class Foo : Thread { int index; this(int i, ?delegate?) { super(?delegate?); index = i; } ~this() { allDone.countDown(); } // or maybe put this in the 'stop'(?) type method? } "?delegate?" above as I cannot remember the syntax off hand. ReganI'll report back after some more experimenting...
Apr 04 2005
+| class Index { int n; this(int n) { this.n = n; } }| module test; | import std.thread; | import std.stdio; | import locks.countdown;+| Index ix = new Index(i);| | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | +| for (int i=0; i < 5 ;i++) {+| int test = 99;| t[i] = new Thread( | delegate int() { +| printf(" thread %d\n",ix.n); | allDone.countDown(); | return 0; +| });So they all ref the same (last one) new'ed Index. The ref apparently is still sitting on the stack in main()'s frame, even though "test" should have stomped on it as I understand D's interspersed allocation method - apparently the internal blocks get their own frame above the main() one, and they still are there, but there is space reserved in main ()'s for the "test" allocation.| | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[j].start(); | | allDone.wait(); | printf("All done.\n"); |} +++++++ Now prints: |Starting threads. | thread 4 | thread 4 | thread 4 | thread 4 | thread 4 |All done.New try with results below: |// dmd test -L locks.lib |module test; |import std.thread; |import std.stdio; |import std.c.time; |import locks.countdown; | ||class Indexed { | | private int n; | private Thread t; | private CountDownLatch* cdlp; | | this ( int n, CountDownLatch* cdlp ) | { | this.n = n; | this.cdlp = cdlp; | this.t = new Thread(delegate int() { | printf(" thread %d\n",n); | cdlp.countDown(); /* (Access Violation */ | return 0; | } | ); | } | void run () { t.start(); } |} | |void main ( char[][] args ) |{ | CountDownLatch allDone = new CountDownLatch(5); | Indexed[5] ixt; | | for (int i=0; i < 5 ;i++) | ixt[i] = new Indexed(i,&allDone); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | ixt[j].run(); | |// allDone.wait(); --> (never ends) | printf("All done.\n"); |} New output is: |Starting threads. | thread 7 |Error: Access Violation | thread 7 |Error: Access Violation | thread 7 |Error: Access Violation | thread 7 |Error: Access Violation | thread 7 |Error: Access Violation |All done. This is just weird - I'm abandoning this path. See below:How do I best do what I obviously want to? I can thnk of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack?From Ben Hinkle:Glad to see someone using the Locks lib. When I was testing things out I ran into the same conundrum you did. A simple way (eg not very generic but it gets the job done) of solving the problem is to loop over t looking for the thread in question: t[i] = new Thread( delegate int() { int n; for(n=0;n<5;n++) { if (Thread.getThis() is t[n]) break; } printf(" thread %d\n",n); allDone.countDown(); return 0; });I am trying to pass a work index into the thread, and this will probably solve that part. It works for me as you coded above. Thanks for all the mintl stuff, especially the concurrent and locks. I don't understand the trouble I had with CountDownLatch - I haven't investigated it yet, but it does work in your method above. I still have some problems in my multithreaded multiple ping, but I think I'm getting independent thread activity now. -Larry
Apr 04 2005
Try this: import std.c.windows.windows; import std.thread; import std.stdio; class RThread : Thread { int index; this(int i) { super(&run); index = i; } int run() { writefln("thread %d",index); //allDone.countDown(); return 0; } } int main(char[][] args) { //CountDownLatch allDone = new CountDownLatch(5); RThread[5] threads; for (int i = 0; i < threads.length; i++) threads[i] = new RThread(i); foreach(RThread t; threads) t.start(); //allDone.wait(); Sleep(10000); printf("All done.\n"); return 0; } Note: I dont have the concurrent stuff so I have included it but commented it out. Regan
Apr 04 2005
Alternate solution using seperate object, not derived from Thread: import std.c.windows.windows; import std.thread; import std.stdio; class Other { int index; this(int i) { index = i; } int run() { writefln("thread %d",index); //allDone.countDown(); return 0; } } int main(char[][] args) { //CountDownLatch allDone = new CountDownLatch(5); Thread[5] threads; Other o; for (int i = 0; i < threads.length; i++) { o = new Other(i); threads[i] = new Thread(&o.run); } foreach(Thread t; threads) t.start(); //allDone.wait(); Sleep(10000); printf("All done.\n"); return 0; } Regan
Apr 04 2005
Ooops. One important correction is that the 'allDone' variable actually needs to be global, or otherwise accessable from the RThread or Other classes. eg. import ..etc.. //global variable/singleton CountDownLatch allDone; //module static ctor, executed when program starts static this() { allDone = new CountDownLatch(5); } class RThread : Thread { ..etc.. } class Other { ..etc.. } int main(char[][] args) { ..etc.. } Regan
Apr 04 2005
Corrected typing error in code... In article <d2sdsg$9ca$1 digitaldaemon.com>, Larry Cowan says...| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[j].start(); | | allDone.wait(); | printf("All done.\n"); |} prints: |Starting threads. | thread 5 | thread 5 | thread 5 | thread 5 | thread 5 |All done. How do I best do what I obviously want to? I can thnk of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack? Do I have to make one for them on the heap to actually get independent activity, then have them pick up an index from a locked counter?
Apr 04 2005
Glad to see someone using the Locks lib. When I was testing things out I ran into the same conundrum you did. A simple way (eg not very generic but it gets the job done) of solving the problem is to loop over t looking for the thread in question:| t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | });t[i] = new Thread( delegate int() { int n; for(n=0;n<5;n++) { if (Thread.getThis() is t[n]) break; } printf(" thread %d\n",n); allDone.countDown(); return 0; });
Apr 04 2005
I should add that if you are on Linux and you start seeing seg-v's on program exit that is because D doesn't wait for all the threads to finish before running gc_term and exiting. So basically you need be very careful before you return from your main() to make sure the threads are completely done and - not just by waiting for a latch, either, since the threads you create still need to execute for a bit after they decrement the latch count. Mike Swieton and I had some posts ages ago about proposals to change how D exits w.r.t. waiting for child threads but nothing changed. Just thought I'd mention it because it can be very confusing trying to debug that behavior. See for example http://www.digitalmars.com/d/archives/digitalmars/D/bugs/156.html "Larry Cowan" <Larry_member pathlink.com> wrote in message news:d2sdsg$9ca$1 digitaldaemon.com...| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[i].start(); | | allDone.wait(); | printf("All done.\n"); |}
Apr 04 2005
Larry Cowan wrote:| module test; | import std.thread; | import std.stdio; | import locks.countdown; | | void main ( char[][] args ) | { | CountDownLatch allDone = new CountDownLatch(5); | Thread[5] t; | | for (int i=0; i < 5 ;i++) | t[i] = new Thread( | delegate int() { | printf(" thread %d\n",i); | allDone.countDown(); | return 0; | }); | | printf("Starting threads.\n"); | | for (int j=0; j < 5 ;j++) | t[i].start(); | | allDone.wait(); | printf("All done.\n"); |} prints: |Starting threads. | thread 5 | thread 5 | thread 5 | thread 5 | thread 5 |All done. How do I best do what I obviously want to? I can thnk of ways, but they are all somewhat convoluted and may be susceptible to similar problems. Do the delegates actually have their own (section of the?) stack? Do I have to make one for them on the heap to actually get independent activity, then have them pick up an index from a locked counter?The problem is, you're asking the threads to print the value of "i", but by the time they've started, "i" doesn't have the value it used to have, so that's why they're all printing 5. Extending Thread (as Regan said) is the way I've done things like this. -- Carlos Santander Bernal JP2, you'll always live in our minds
Apr 04 2005