查看Linux进程CPU过高具体的线程堆栈(不中断程序)
转自:http://blog.csdn.net/mergerly/article/details/47731305
1、TOP命令,找到占用CPU最高的进程
- $ top
- top - 20:11:45 up 850 days, 1:18, 3 users, load average: 1.04, 1.01, 0.99
- Tasks: 61 total, 1 running, 60 sleeping, 0 stopped, 0 zombie
- Cpu(s): 1.4% us, 0.1% sy, 0.0% ni, 98.3% id, 0.1% wa, 0.0% hi, 0.2% si
- Mem: 16418172k total, 15693376k used, 724796k free, 1146696k buffers
- Swap: 10223608k total, 0k used, 10223608k free, 12537692k cached
- PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
- 24714 ztgame 16 0 1409m 1.2g 4252 S 99.6 7.4 2390:57 IMVChannelServe
- 1 root 16 0 4772 520 432 S 0.0 0.0 0:03.43 init
- 2 root RT 0 0 0 0 S 0.0 0.0 0:05.75 migration/0
- 3 root 34 19 0 0 0 S 0.0 0.0 5:22.97 ksoftirqd/0
- 4 root RT 0 0 0 0 S 0.0 0.0 0:07.90 migration/1
- 5 root 34 19 0 0 0 S 0.0 0.0 0:00.27 ksoftirqd/1
- 6 root RT 0 0 0 0 S 0.0 0.0 0:04.07 migration/2
- 7 root 34 19 0 0 0 S 0.0 0.0 0:00.47 ksoftirqd/2
- 8 root RT 0 0 0 0 S 0.0 0.0 0:04.00 migration/3
- 9 root 34 19 0 0 0 S 0.0 0.0 0:00.33 ksoftirqd/3
2、通过TOP -H -p 进程ID,找到具体的线程占用情况,Shift+H可以开启关闭线程显示
- $ top -H -p 24714
- top - 20:15:30 up 850 days, 1:22, 3 users, load average: 1.26, 1.09, 1.02
- Tasks: 16 total, 1 running, 15 sleeping, 0 stopped, 0 zombie
- Cpu(s): 24.8% us, 0.3% sy, 0.0% ni, 73.1% id, 0.0% wa, 0.0% hi, 1.8% si
- Mem: 16418172k total, 15701376k used, 716796k free, 1146704k buffers
- Swap: 10223608k total, 0k used, 10223608k free, 12546048k cached
- PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
- 24729 ztgame 16 0 1409m 1.2g 4252 R 97.0 7.4 2307:22 IMVChannelServe
- 24721 ztgame 15 0 1409m 1.2g 4252 S 2.0 7.4 84:22.40 IMVChannelServe
- 24714 ztgame 16 0 1409m 1.2g 4252 S 0.0 7.4 0:03.80 IMVChannelServe
- 24716 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.00 IMVChannelServe
- 24717 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.04 IMVChannelServe
- 24718 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.00 IMVChannelServe
- 24719 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.08 IMVChannelServe
- 24720 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.01 IMVChannelServe
- 24722 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.00 IMVChannelServe
- 24723 ztgame 16 0 1409m 1.2g 4252 S 0.0 7.4 0:00.00 IMVChannelServe
- 24724 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.01 IMVChannelServe
- 24725 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:09.83 IMVChannelServe
- 24726 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.00 IMVChannelServe
- 24727 ztgame 15 0 1409m 1.2g 4252 S 0.0 7.4 0:00.76 IMVChannelServe
- 24728 ztgame 16 0 1409m 1.2g 4252 S 0.0 7.4 0:00.53 IMVChannelServe
- 24730 ztgame 16 0 1409m 1.2g 4252 S 0.0 7.4 2:42.18 IMVChannelServe
3、通过命令pstack 进程ID显示线程堆栈,LWP 24729对应线程ID的堆栈,就是占用CPU最高的堆栈,可以具体分析什么原因造成的。
- $ pstack 24714
- Thread 16 (Thread 1084229984 (LWP 24716)):
- #0 0x00000039c5a901d5 in __nanosleep_nocancel () from /lib64/tls/libc.so.6
- #1 0x00000039c5ac3058 in usleep () from /lib64/tls/libc.so.6
- #2 0x00000000005ebe10 in zVerifyThread::run ()
- #3 0x00000000005e9d29 in zThread::threadFunc ()
- #4 0x00000039c6106137 in start_thread () from /lib64/tls/libpthread.so.0
- #5 0x00000039c5ac9883 in clone () from /lib64/tls/libc.so.6
- Thread 15 (Thread 1094719840 (LWP 24717)):
- #0 0x00000039c5a901d5 in __nanosleep_nocancel () from /lib64/tls/libc.so.6
- #1 0x00000039c5ac3058 in usleep () from /lib64/tls/libc.so.6
- #2 0x00000000005ebe10 in zVerifyThread::run ()
- #3 0x00000000005e9d29 in zThread::threadFunc ()
- #4 0x00000039c6106137 in start_thread () from /lib64/tls/libpthread.so.0
- #5 0x00000039c5ac9883 in clone () from /lib64/tls/libc.so.6
- Thread 14 (Thread 1105209696 (LWP 24718)):
- #0 0x00000039c5a901d5 in __nanosleep_nocancel () from /lib64/tls/libc.so.6
- #1 0x00000039c5ac3058 in usleep () from /lib64/tls/libc.so.6
- #2 0x00000000005ebe10 in zVerifyThread::run ()
- #3 0x00000000005e9d29 in zThread::threadFunc ()
- #4 0x00000039c6106137 in start_thread () from /lib64/tls/libpthread.so.0
- #5 0x00000039c5ac9883 in clone () from /lib64/tls/libc.so.6
- Thread 13 (Thread 1115699552 (LWP 24719)):
- #0 0x00000039c5a901d5 in __nanosleep_nocancel () from /lib64/tls/libc.so.6
- #1 0x00000039c5ac3058 in usleep () from /lib64/tls/libc.so.6
- #2 0x00000000005ebe10 in zVerifyThread::run ()
- #3 0x00000000005e9d29 in zThread::threadFunc ()
- #4 0x00000039c6106137 in start_thread () from /lib64/tls/libpthread.so.0
- #5 0x00000039c5ac9883 in clone () from /lib64/tls/libc.so.6
- Thread 3 (Thread 1220598112 (LWP 24729)):
- #0 0x00000039c5a71e87 in memset () from /lib64/tls/libc.so.6
- #1 0x00000000004fa591 in ChannelTask::forwardToClientByMedia ()
- #2 0x0000000000506220 in ChannelTask::parseClientMsg_Normal ()
- #3 0x000000000051ef55 in ChannelTask::parseClientMsg ()
- #4 0x000000000051f070 in ChannelTask::cmdMsgParse_Forward ()
- #5 0x000000000051f1d1 in ChannelTask::cmdMsgParse ()
- #6 0x000000000051f414 in ChannelTask::processCmd ()
- #7 0x0000000000523ea8 in ChannelTaskManager::processCmd ()
- #8 0x0000000000525ddd in ChannelTimeTick::run ()
- #9 0x00000000005e9d29 in zThread::threadFunc ()
- #10 0x00000039c6106137 in start_thread () from /lib64/tls/libpthread.so.0
- #11 0x00000039c5ac9883 in clone () from /lib64/tls/libc.so.6
- Thread 2 (Thread 1231087968 (LWP 24730)):
- #0 0x00000039c610af8b in __lll_mutex_lock_wait ()
- #1 0x0000000000000001 in ?? ()
- #2 0x0000000000000065 in ?? ()
- #3 0x00000039c6107d87 in pthread_mutex_lock () from /lib64/tls/libpthread.so.0
- #4 0x0000003a500ae29e in operator delete () from /usr/lib64/libstdc++.so.6
- #5 0x000000000053f59d in ChannelLoadClient::processCmd_DB ()
- #6 0x00000000005986c9 in GameAppClient::processTaskCmd_DB ()
- #7 0x00000039c5a901e3 in __nanosleep_nocancel () from /lib64/tls/libc.so.6
- #8 0x0000000000000000 in ?? ()
- Thread 1 (Thread 182894183104 (LWP 24714)):
- #0 0x00000039c5ac9c5c in epoll_wait () from /lib64/tls/libc.so.6
- #1 0x0000000000620cac in zTCPServer::accept ()
- #2 0x00000000005f9c0d in zNetService::serviceCallback ()
- #3 0x00000000005f89e3 in zService::main ()
- #4 0x0000000000564298 in main ()
查看Linux进程CPU过高具体的线程堆栈(不中断程序)的更多相关文章
- 【转载】查看Linux进程CPU过高具体的线程堆栈(不中断程序)
具体的命令经常忘记,毕竟用的不是很多.为了避免去找备份一下 1.TOP命令,找到占用CPU最高的进程 $ top top - 20:11:45 up 850 days, 1:18, 3 users ...
- linux ps命令,查看某进程cpu和内存占用率情况, linux ps命令,查看进程cpu和内存占用率排序。 不指定
背景:有时需要单看某个进程的CPU及占用情况,有时需要看整体进程的一个占用情况.一. linux ps命令,查看某进程cpu和内存占用率情况[root@test vhost]# ps auxUSER ...
- Java进程CPU使用率高排查
Java进程CPU使用率高排查 生产java应用,CPU使用率一直很高,经常达到100%,通过以下步骤完美解决,分享一下.1.jps 获取Java进程的PID.2.jstack pid >> ...
- 查看linux服务器CPU相关
查看linux服务器CPU相关: 1.查看物理CPU个数 cat /proc/cpuinfo| grep "physical id"| sort| uniq| wc -l 2.查看 ...
- 查看Linux服务器CPU总核数
下面介绍查看Linux服务器CPU总核数的方法. 通过/proc/cpuinfo可查看CPU个数及总核数. [root@kevin ~]# grep processor /proc/cpuinfo | ...
- linux查看某个进程CPU消耗较高的具体线程或程序的方法
目前我们的监控,可以发现消耗较高CPU的进程(阀值为3个CPU),通过监控我们可以找到消耗较高CPU的进程号: 通过进程号pid,我们在linux上可以通过top –H –p <pid> ...
- java进程CPU飙高
因为这段时间一直在弄监控,但是工作还是在进行中 因为机器不多,所以今天早上巡检了一下,看到一台生产机器上的CPU飙高 top
- JVM进程cpu飙高分析
在项目快速迭代中版本发布频繁 近期上线报错一个JVM导致服务器cpu飙高 但内存充足的原因现象. 对于耗内存的JVM程序来而言, 基本可以断定是线程僵死(死锁.死循环等)问题. 这里是纪录一下排 ...
- JVM 之 Linux定位CPU过高问题及优化
项目部署以后出行卡顿现象,所以对问题进行了排查,记录一下排查过程 (从CSDN编辑器贴过来的,图有水印) 1.找进程 top 可以发现,是Java进程导致的CPU过高,致使系统卡顿 2.找线程 ps ...
随机推荐
- PHP 字符串截取()[]{} 中内容
$str="你好<我>(爱)[北京]{天安门}"; echo f1($str); //返回你好 echo f2($str); //返回我 echo f3($str); ...
- 2018-2019-2 网络对抗技术 20165301 Exp5 MSF基础应用
2018-2019-2 网络对抗技术 20165301 Exp5 MSF基础应用 实践原理 1.MSF攻击方法 主动攻击:扫描主机漏洞,进行攻击 攻击浏览器 攻击其他客户端 2.MSF的六个模块 查看 ...
- 《高性能MySQL》学习笔记
第1章 MySQL架构与历史 1.2 并发控制 MySQL在两个层面实现并发控制:服务器层与存储引擎层. 读锁和写锁: 在处理并发读或写时,可以通过实现一个由两种锁组成的系统来解决问题. 这两种锁通常 ...
- 【PAT】1103 Integer Factorization(30 分)
The K−P factorization of a positive integer N is to write N as the sum of the P-th power of K positi ...
- js导出带格式的表格(包括单元格合并,字体样式等)
function HtmlExportToExcelForEntire() { var uri = 'data:application/vnd.ms-excel;base64,', template ...
- 007.KVM虚机时间-快照管理
一 快照管理 1.1 创建快照 [root@kvm-host ~]# virsh snapshot-create vm03-centos6.8 [root@kvm-host ~]# virsh sna ...
- SpringMVC(十) RequestMapping RequestHeader注解
在控制器方法中,通过类似 RequestHeader(value="Accept-Lanaguage") String lan 可以获取请求头信息. 控制器方法: package ...
- hdu 5203 && BC Round #37 1002
代码参考自:xyz111 题意: 众所周知,萌萌哒六花不擅长数学,所以勇太给了她一些数学问题做练习,其中有一道是这样的:勇太有一根长度为n的木棍,这个木棍是由n个长度为1的小木棍拼接而成,当然由于时间 ...
- Linux shell 脚本小记2
.从文件读取 while read line do echo "line=$line" done < file.txt .将字符串转换为数组,并进行遍历 str=" ...
- 14、Redis的复制
写在前面的话:读书破万卷,编码如有神 --------------------------------------------------------------------------------- ...