I think this is best dealt with by the OS. On Wed, Sep 9, 2009 at 9:40 PM, <jon_zhou@agilent.com> wrote:
Hi there:
I think the "sched_yield" will put itself to the end of the waiting queue If there is no other waiting task, the program will be scheduled again soon.
Which is the better way to receive msg?
#1 While (1) { Mq.receive(...) }
Or #2 While (1) { if (!Mq.try_receive()) sleep(xx) }
I think the former can get the best throughput performance but the latter can get the less resource used.
Any comments?
Thanks Jon
-----Original Message----- From: ZHOU,JON (A-China,ex2) Sent: Tuesday, September 08, 2009 9:55 PM To: ZHOU,JON (A-China,ex2); 'boost@lists.boost.org' Subject: RE: [interprocess]why message_queue receive() will consume so much CPU resource
The code can trap in sched_yield (),(see below)
why it still consume high CPU resource?
(gdb) bt #0 0xffffe410 in __kernel_vsyscall () #1 0x401b706c in sched_yield () from /lib/tls/libc.so.6 #2 0x0804e88f in thread_yield () at ../boost138inst/include/boost-1_38/boost/interprocess/detail/os_thread_functions.hpp:1 04 #3 0x0804c22b in boost::interprocess::interprocess_condition::do_timed_wait ( this=0x40018018, tout_enabled=false, abs_time=@0xbff77a70, mut=@0x40018014) at ../boost138inst/include/boost-1_38/boost/interprocess/sync/emulation/interprocess_cond ition.hpp:125 #4 0x0804bfc2 in boost::interprocess::interprocess_condition::do_wait (this=0x40018018, mut=@0x40018014) at ../boost138inst/include/boost-1_38/boost/interprocess/sync/emulation/interprocess_cond ition.hpp:70
-----Original Message----- From: ZHOU,JON (A-China,ex2) Sent: Tuesday, September 08, 2009 10:21 AM To: boost@lists.boost.org; ZHOU,JON (A-China,ex2) Subject: [interprocess]why message_queue receive() will consume so much CPU resource
Hi there
I just found that message_queue::receive() or message_queue:: timed_receive () consume high CPU resource, (99% CPU) the code like the below:
//create message_queue
date d(2009,Sep,8); //an arbitrary date //construct a time by adding up some durations durations ptime t1(d, hours(0)+minutes(0)+seconds(1)+millisec(0));
while (1) { for(int i = 0; i < 100; ++i){ int number; //mq.receive(&number, sizeof(number), recvd_size, priority); mq.timed_receive(&number, sizeof(number), recvd_size, priority,t1); } }
As I know, both the receive() & timed_receive() are block method, there should be sleep mechanism inside. why it still need so much CPU resource?
Thanks jon
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