Linux load average的意思就是说 目前有ready的进程 但是cpu都在满,然后这个数是一分钟的平均值。公式:loadvg = tasks running + tasks waiting (for cores) + tasks blocked. 让CPU满是最简单的方式
为了更好理解load average,Tim某天在群征集一段代码使load average最高。于是乎,群友纷纷出手。
Erlang版
[~]$ cat load10 #!/usr/bin/env escript %% -- erlang -- %%! -smp enable -sname load10 main(_) -> I = erlang:system_info(scheduler_id), random(5). random(I) when I > 10 -> random(I+1); random(I) -> spawn(fun() -> random(I+1) end), random(I+1).
此代码将Linux load跑到了10-20,取得了领先
Shell版
#!/bin/sh for((i=0;i<10;i++));do { for((j=0;j<1000000000000;j++));do echo '1'>>1; done }& done
Ruby
ruby -e "require 'thread';t = Mutex.new;10.times {fork {10000000.times{t.synchronize {x=Time.now}}}}"
Java版
import java.io.*; public class LoadTest { public static void main(String[] args) throws IOException, InterruptedException { int count = 1000; for (int i = 0; i < count; i++) { final Thread t = new Thread(new Runnable() { public void run() { try { while (true) { RandomAccessFile file = new RandomAccessFile("/tmp/test.bin", "rw"); file.seek(1024 * 1024 * 500); file.write(1); file.close(); } } catch (IOException e) { } } }); t.start(); } Thread.currentThread().join(); } }
C语言版
void main () { int i=0; for (i = 0; i < 1000; i++) { if (fork () > 0) { continue; } while (1) ; } getchar(); }
此版本远远超过了上面的Erlang版本,将load拉到1000+
Java版本2
public class Test { public static void main(String[] args) { for (int i = 0; i < 1000; i++) { new Thread() { @Override public void run() { while (true) ; } }.start(); } } }
此版本和上面C版本效果类似
部分讨论:
不对呀,这个方法以前不行
CentOS 6吧[惊讶]
CentOS 5
有几个版本有bug,cpu的hz拿的不对,load特别低,7就好了[坏笑]
Java版本3
写两类线程,一类进行lockObject wait, 一类进行lockObject notifyall。 cpu上下文切换消耗 竞争激烈。Load能很高。
public class LoadHighDemo { public static void main(String[] args) throws Exception { LoadHighDemo demo = new LoadHighDemo(); demo.runTest(); } private void runTest() throws Exception { Object[] locks = new Object[5000]; for (int i = 0; i < 5000; i++) { locks[i] = new Object(); new Thread(new WaitTask(locks[i])).start(); new Thread(new NotifyTask(locks[i])).start(); } } class WaitTask implements Runnable { private Object lockObject = null; public WaitTask(Object obj) { lockObject = obj; } public void run() { while (true) { try { synchronized (lockObject) { lockObject.wait(new java.util.Random().nextInt(10)); } } catch (Exception e) {; } } } } class NotifyTask implements Runnable { private Object lockObject = null; public NotifyTask(Object obj) { lockObject = obj; } public void run() { while (true) { synchronized (lockObject) { lockObject.notifyAll(); } try { Thread.sleep(new java.util.Random().nextInt(5)); } catch (InterruptedException e) { } } } } }
上物理机把5000个线程改成更多,如50000也行。
直接while true改5000应该能更多……
我的应该能一直上涨, load 1000, 2000, 3000
可以试试parkNanos(1)
刚才卡了。。。
看来不能在本机上跑
Python版
把5分钟 load avg 稳定在1024了
在本群感到了一种情怀。。。
#!/usr/bin/env python import os import sys import time def load_add_1(): time.sleep(30) fd=os.open("test.txt",os.O_CREAT|os.O_RDWR|os.O_SYNC, 0644) for i in xrange(10000*100): os.write(fd," "*100) sys.exit(0) for i in xrange(8192): if os.fork() == 0: load_add_1()
此版本将load跑到8192
C语言版本2
https://gist.github.com/hongqn/37cdfde04d0c5a03fede
#include <unistd.h> #include <stdio.h> #define LOAD 16384 int main() { int i; char s[256]; for (i=0; i<LOAD; i++) { if (vfork()) { return 0; } } scanf("%s", s); return 0; }
macbook 上 virtualbox 里一个 2G mem 单 core cpu 的虚机就可以轻松达到16,000
这个的原理是通过 vfork 产生指定个数的 D 状态进程,从而提高 loadVfork() differs from fork in that the child borrows the parent’s memory and thread of control until a call to execve(2) or an exit (either by a call to exit(2) or abnormally.) The parent process is suspended while the child is using its resources.
vfork 的子进程只要不 execve 或者退出,父进程就一直挂着(在D状态)。这里就是让最后一个子进程用 scanf 等输入
通过以上题目,你是看到了热闹,还是看到了门道?
c 语言版本二中, if(){return } 这个不是会把程序挂掉么?
贡献一个
#/bin/bash
yes
楼上不行吧,死循环一直输出y负载肯定上不去。
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