soft and program: comp.programming.threads - 11 new messages in 7 topics

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* Hopscotch Hashing - new version - 1 messages, 1 author
http://groups.google.com/group/comp.programming.threads/t/cb69d4304c75f463?hl=en
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* Air Jordans 16 (paypal payment)( www.sneaker-shop08 - 1 messages, 1 author
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* invalid conversion from 'void* (*)()' to 'void* (*)(void*)' when using
pthread in C++ - 5 messages, 4 authors
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==============================================================================
TOPIC: Hopscotch Hashing - new version
http://groups.google.com/group/comp.programming.threads/t/cb69d4304c75f463?hl=en
==============================================================================

== 1 of 1 ==
Date: Thurs, Mar 26 2009 3:12 am
From: Dmitriy Vyukov

On Mar 25, 5:55 pm, "Chris M. Thomasson" <n...@spam.invalid> wrote:

> Right. Also, even on SPARC TSO, some loads in the critical-section can be
> hoisted up above the CAS in the lock function of TTASLock. AFAICT, the CAS
> (e.g., CASIO) is naked in their implementation. IIRC, #StoreLoad is the only
> membar that is not a NOP on TSO SPARC. For instance, you need it for
> Petersons Algorihtm, even if you use atomic RMW. This differs from Intel
> where a LOCK instruction acts like a #StoreLoad for WB memory.

Here is one moment I don't quite understand.
CAS includes both load and store. Let's assume that there is no
#StoreLoad after CAS, only #LoadLoad | #LoadStore (i.e. load-acquire).
Here is pseudo code for mutex unlock followed by mutex lock:

// some code from previous critical section:
tmp = data;
data = tmp + 1;
// mutex unlock:
membar #StoreStore | #LoadStore;
lock = UNLOCKED;

// mutex lock (unrolled CAS):
tmp = lock;
if (tmp != UNLOCKED) goto retry;
lock = LOCKED;
membar #LoadLoad | #LoadStore;
// critical section:
tmp = data;
data = tmp + 1;

At maximum this can be reordered as:

// mutex lock (unrolled CAS):
tmp = lock;
if (tmp != UNLOCKED) goto retry;
membar #LoadLoad | #LoadStore;
// critical section:
tmp = data;
data = tmp + 1;
lock = LOCKED;

The question is: can this reordering lead to some problem?

Code from previous critical section and from current critical section
can not be interleaved (as I understand), because load that yield
UNLOCKED can not hoist above store of UNLOCKED.

Consider following simplified "one-shot" mutex:

initially lock = LOCKED

thread 1:
tmp = data;
data = tmp + 1;
membar #StoreStore | #LoadStore;
lock = UNLOCKED;

thread 2:
if (lock == UNLOCKED)
{
membar #LoadLoad | #LoadStore;
tmp = data;
data = tmp + 1;
}

Here #LoadLoad | #LoadStore is enough in thread 2, because there is
just no stores to lock in thread 2. So why #LoadLoad | #LoadStore is
not enough in general mutex lock() implementation?

--
Dmitriy V'jukov

==============================================================================
TOPIC: Bluetooth Sunglasses
http://groups.google.com/group/comp.programming.threads/t/0072f598f7f227e8?hl=en
==============================================================================

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TOPIC: Air Jordans 16 (paypal payment)( www.sneaker-shop08
http://groups.google.com/group/comp.programming.threads/t/e881baef7d0e10d5?hl=en
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==============================================================================
TOPIC: invalid conversion from 'void* (*)()' to 'void* (*)(void*)' when using
pthread in C++
http://groups.google.com/group/comp.programming.threads/t/8fddb216a36daa87?hl=en
==============================================================================

== 1 of 5 ==
Date: Fri, Mar 27 2009 9:19 am
From: Tim

Hi,
I am using pthread in my main.cpp file and got error in the following
code. Thank you very much for your help!

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>

pthread_mutex_t count_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_mutex_t condition_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t condition_cond = PTHREAD_COND_INITIALIZER;

void *functionCount1();
void *functionCount2();
int count = 0;
#define COUNT_DONE 10
#define COUNT_HALT1 3
#define COUNT_HALT2 6

int main()
{
pthread_t thread1, thread2;

pthread_create( &thread1, NULL, &functionCount1, NULL);
pthread_create( &thread2, NULL, &functionCount2, NULL);
pthread_join( thread1, NULL);
pthread_join( thread2, NULL);

return 0;
}

void *functionCount1()
{
for(;;)
{
pthread_mutex_lock( &condition_mutex );
while( count >= COUNT_HALT1 && count <= COUNT_HALT2 )
{
pthread_cond_wait( &condition_cond, &condition_mutex );
}
pthread_mutex_unlock( &condition_mutex );

pthread_mutex_lock( &count_mutex );
count++;
printf("Counter value functionCount1: %d\n",count);
pthread_mutex_unlock( &count_mutex );

if(count >= COUNT_DONE) return(NULL);
}
}

void *functionCount2()
{
for(;;)
{
pthread_mutex_lock( &condition_mutex );
if( count < COUNT_HALT1 || count > COUNT_HALT2 )
{
pthread_cond_signal( &condition_cond );
}
pthread_mutex_unlock( &condition_mutex );

pthread_mutex_lock( &count_mutex );
count++;
printf("Counter value functionCount2: %d\n",count);
pthread_mutex_unlock( &count_mutex );

if(count >= COUNT_DONE) return(NULL);
}

}

g++ gives error:

== 2 of 5 ==
Date: Fri, Mar 27 2009 11:22 am
From: Torsten Robitzki

Hi Tim,

change
> void *functionCount1();
> void *functionCount2();

void *functionCount1(void*);
void *functionCount2(void*);

best regards
Torsten

--
kostenlose Wirtschaftssimulation: http://www.financial-rumors.de

== 3 of 5 ==
Date: Fri, Mar 27 2009 11:57 am
From: Steve Watt

In article <511bdaf3-eb07-44c4-8e01-eac2bf668bcc@r28g2000vbp.googlegroups.com>,
Tim <timlee126@yahoo.com> wrote:
>Hi,
>I am using pthread in my main.cpp file and got error in the following
>code. Thank you very much for your help!

[ ... ]

>void *functionCount1();
>void *functionCount2();

These prototypes don't agree with

> pthread_create( &thread1, NULL, &functionCount1, NULL);
> pthread_create( &thread2, NULL, &functionCount2, NULL);

the argument type that is expected to pthread_create().

The pthread_* functions are C language. The C++ prototype for
a thread function is

extern "C" void *thread_function(void *);

Enjoy,
--
Steve Watt KD6GGD PP-ASEL-IA ICBM: 121W 56' 57.5" / 37N 20' 15.3"
Internet: steve @ Watt.COM Whois: SW32-ARIN
Free time? There's no such thing. It just comes in varying prices...

== 4 of 5 ==
Date: Fri, Mar 27 2009 3:10 pm
From: Tim

Thanks Torsten and Steve. You solved my problem!

== 5 of 5 ==
Date: Fri, Mar 27 2009 3:50 pm
From: David Schwartz

On Mar 27, 9:19 am, Tim <timlee...@yahoo.com> wrote:

Others have pointed out the mistake that caused you to post. However,
you have a very serious misunderstanding about how to use condition
variables:

> void *functionCount1()
> {
> for(;;)
> {
> pthread_mutex_lock( &condition_mutex );
> while( count >= COUNT_HALT1 && count <= COUNT_HALT2 )
> {
> pthread_cond_wait( &condition_cond, &condition_mutex );
> }

The point of the 'while' loop is to keep looping until you know the
condition is true. You check it again when 'pthread_cond_wait' returns
because just because it was true in the past doesn't mean it's true in
the future.

The whole point of this code is to make absolutely sure that the
following code does not execute unless the condition is true.

> pthread_mutex_unlock( &condition_mutex );

Whoops! You just unlocked the mutex. Now the condition may no longer
be true. This completely defeats the point of the 'while' loop as the
code below here may run whether or not the condition is true.

> pthread_mutex_lock( &count_mutex );
> count++;
> printf("Counter value functionCount1: %d\n",count);
> pthread_mutex_unlock( &count_mutex );
>
> if(count >= COUNT_DONE) return(NULL);
> }
> }

Why do you have two mutexes, both protecting the count?

And now you can see your 'while' loop is busted too:

> void *functionCount2()
> {
> for(;;)
> {
> pthread_mutex_lock( &condition_mutex );
> if( count < COUNT_HALT1 || count > COUNT_HALT2 )
> {
> pthread_cond_signal( &condition_cond );
> }
> pthread_mutex_unlock( &condition_mutex );
>
> pthread_mutex_lock( &count_mutex );
> count++;

Here, you modify 'count' without holding the condition_mutex. But the
other thread *reads* 'count' holding only the condition mutex. Boom.

One thread may not modify a variable while another thread is or may be
reading it. This thread may modify 'count' while the first thread is
reading it in the 'while' loop.

> printf("Counter value functionCount2: %d\n",count);
> pthread_mutex_unlock( &count_mutex );
>
> if(count >= COUNT_DONE) return(NULL);
> }
>
> }

You must use the same mutex to protect the predicate as you acquire/
release with the condition variable. Otherwise, the code makes no
sense.

==============================================================================
TOPIC: www.stefsclothes.com Wholesale Nike AirMax,nike shox,Air Jordan,air
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http://groups.google.com/group/comp.programming.threads/t/96f000ef72a432c7?hl=en
==============================================================================

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TOPIC: Hogan shoes (paypal payment)( www.niketrade08.cn )
http://groups.google.com/group/comp.programming.threads/t/d3f275105e08f72d?hl=en
==============================================================================

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