Shell，运行四个并行进程

目前我被困在运行耗时的模拟有效。目的是并行运行4个仿真，因为它是一个单线程应用程序和一个四核系统。我必须变种shell脚本：Shell，运行四个并行进程

./sim -r 1 & 
./sim -r 2 & 
./sim -r 3 & 
./sim -r 4 & 
wait 
./sim -r 5 & 
./sim -r 6 & 
./sim -r 7 & 
./sim -r 8 & 
wait 
... (another 112 jobs)

有了这段代码，就有了一次又一次的等待。我也尝试将这些任务分成四个脚本并运行，结果是一个脚本完成了，另一个脚本剩下约30％的工作。我无法预测模拟需要多长时间。

任何建议有4个模拟随时运行？

来源

2013-11-02 Daniel

有趣的问题。 AFAIK，它不能在'bash'（或Korn shell，Bourne shell或POSIX shell）中完成。您可以等待一个特定的进程（'bash'）或所有进程（所有进程），但不能同时处理一组进程中的一个进程。我可能会使用Perl（或者Python）来运行;它可以等待任何进程死亡，然后启动另一个进程。 –

安装moreutils包在Ubuntu，然后使用parallel实用程序：

parallel -j 4 ./sim -r -- 1 2 3 4 5 6 7 8 ...

来源

2013-11-02 22:33:25 pobrelkey

如果你不想安装parallel实用程序（它看起来整洁假设它的工作原理所示），那么你可以适应此Perl脚本（基本上，改变所执行的命令），并可能降低监测：

#!/usr/bin/env perl 
use strict; 
use warnings; 
use constant MAX_KIDS => 4; 

$| = 1; 

my %pids; 
my $kids = 0; # Number of kids 

for my $i (1..20) 
{ 
    my $pid; 
    if (($pid = fork()) == 0) 
    { 
     my $tm = int(rand() * (10 - 2) + 2); 
     print "sleep $tm\n"; 
     # Using exec in a block on its own is the documented way to 
     # avoid the warning: 
     # Statement unlikely to be reached at filename.pl line NN. 
     # (Maybe you meant system() when you said exec()?) 
     # Yes, I know the print and exit statements should never be 
     # reached, but, dammit, sometimes things go wrong! 
     { exec "sleep", $tm; } 
     print STDERR "Oops: couldn't sleep $tm!\n"; 
     exit 1; 
    } 
    $pids{$pid} = 1; 
    $kids++; 
    my $time = time; 
    print "PID: $pid; Kids: $kids; Time: $time\n"; 
    if ($kids >= MAX_KIDS) 
    { 
     my $kid = waitpid(-1, 0); 
     print "Kid: $kid ($?)\n"; 
     if ($kid != -1) 
     { 
      delete $pids{$kid}; 
      $kids--; 
     } 
    } 
} 

while ((my $kid = waitpid(-1, 0)) > 0) 
{ 
    my $time = time; 
    print "Kid: $kid (Status: $?); Time: $time\n"; 
    delete $pids{$kid}; 
    $kids--; 
} 

# This should not do anything - and doesn't (any more!). 
foreach my $pid (keys %pids) 
{ 
    printf "Undead: $pid\n"; 
}

输出示例：

PID: 20152; Kids: 1; Time: 1383436882 
PID: 20153; Kids: 2; Time: 1383436882 
sleep 5 
PID: 20154; Kids: 3; Time: 1383436882 
sleep 7 
sleep 9 
PID: 20155; Kids: 4; Time: 1383436882 
sleep 4 
Kid: 20155 (0) 
PID: 20156; Kids: 4; Time: 1383436886 
sleep 4 
Kid: 20152 (0) 
PID: 20157; Kids: 4; Time: 1383436887 
sleep 2 
Kid: 20153 (0) 
PID: 20158; Kids: 4; Time: 1383436889 
sleep 9 
Kid: 20157 (0) 
PID: 20159; Kids: 4; Time: 1383436889 
sleep 6 
Kid: 20156 (0) 
PID: 20160; Kids: 4; Time: 1383436890 
sleep 6 
Kid: 20154 (0) 
PID: 20161; Kids: 4; Time: 1383436891 
sleep 9 
Kid: 20159 (0) 
PID: 20162; Kids: 4; Time: 1383436895 
sleep 7 
Kid: 20160 (0) 
PID: 20163; Kids: 4; Time: 1383436896 
sleep 9 
Kid: 20158 (0) 
PID: 20164; Kids: 4; Time: 1383436898 
sleep 6 
Kid: 20161 (0) 
PID: 20165; Kids: 4; Time: 1383436900 
sleep 9 
Kid: 20162 (0) 
PID: 20166; Kids: 4; Time: 1383436902 
sleep 9 
Kid: 20164 (0) 
PID: 20167; Kids: 4; Time: 1383436904 
sleep 2 
Kid: 20163 (0) 
PID: 20168; Kids: 4; Time: 1383436905 
sleep 6 
Kid: 20167 (0) 
PID: 20169; Kids: 4; Time: 1383436906 
sleep 9 
Kid: 20165 (0) 
PID: 20170; Kids: 4; Time: 1383436909 
sleep 4 
Kid: 20168 (0) 
PID: 20171; Kids: 4; Time: 1383436911 
Kid: 20166 (0) 
sleep 9 
Kid: 20170 (Status: 0); Time: 1383436913 
Kid: 20169 (Status: 0); Time: 1383436915 
Kid: 20171 (Status: 0); Time: 1383436920

来源

2013-11-03 00:04:49

NUMJOBS=30 
NUMPOOLS=4 

seq 1 "$NUMJOBS" | for p in $(seq 1 $NUMPOOLS); do 
    while read x; do ./sim -r "$x"; done & 
done

的for循环创建后台进程从所述共享标准输入读取开始仿真的池。每个后台进程在模拟运行时都会“阻塞”，然后从seq命令中读取下一个作业编号。

没有for循环，这可能是一个比较容易遵循：

seq 1 "$NUMJOBS" | { 
    while read x; do ./sim -r "$x"; done & 
    while read x; do ./sim -r "$x"; done & 
    while read x; do ./sim -r "$x"; done & 
    while read x; do ./sim -r "$x"; done & 
}

假设sim花费的时间来运行一个不平凡的量，第一while将从其标准输入读取1， 2nd 2等。无论哪种sim先完成，即while循环将从read 5起标准输入;下一个结束将读取6，依此类推。一旦最后一次模拟开始，每个read都会失败，导致循环退出。

来源

2013-11-04 14:03:06 chepner

Shell，运行四个并行进程

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