即比如我有100个可执行文件，互相间没有特别的先后执行关系，如CODE:

复制代码 代码如下:

job_1

job_2

job_2

.....

job_100

想用csh/bash来多线程调用执行。

比如一次开5个线程，那么job_1,2,3,4,5一起先开始，那么其中任何一个线程如果先执行完成，则继续执行下一个没有初执行过的文件，如job_6,7,8....，这样一直以所指定的线程数来执行所有100个文件。

我本来想用 "&" 来放入后台，可是这样我一次可以指定5放入后台，但是无法知道其中任何一个程序何时执行完毕，所以也无法继续执行下一个程序啊!

完美解决方案：

复制代码 代码如下:

-(dearvoid@LinuxEden:Forum)-(~/tmp)-

[$$=6718 $?=0] ; cat job_1

#!/bin/bash

n=$((RANDOM % 5 + 1))

echo "$0 sleeping for $n seconds ..."

sleep $n

echo "$0 exiting ..."

-(dearvoid@LinuxEden:Forum)-(~/tmp)-

[$$=6718 $?=0] ; for ((i = 2; i <= 10; ++i)); do cp job_1 job_$i; done

-(dearvoid@LinuxEden:Forum)-(~/tmp)-

[$$=6718 $?=0] ; cat jobs.sh

#!/bin/bash

nParellel=5

nJobs=10

sJobPattern='./job_%d'

aJobs=()

sNextJob=

for ((iNextJob = 1; iNextJob <= nJobs; )); do

for ((iJob = 0; iJob < nParellel; ++iJob)); do

if [ $iNextJob -gt $nJobs ]; then

break;

fi

if [ ! "${aJobs[iJob]}" ] || ! kill -0 ${aJobs[iJob]} 2> /dev/null; then

printf -v sNextJob "$sJobPattern" $((iNextJob++))

echo "$sNextJob starting ..."

$sNextJob &

aJobs[iJob]=$!

fi

done

sleep .1

done

wait

-(dearvoid@LinuxEden:Forum)-(~/tmp)-

[$$=6718 $?=0] ; ./jobs.sh

./job_1 starting ...

./job_1 sleeping for 3 seconds ...

./job_2 starting ...

./job_2 sleeping for 2 seconds ...

./job_3 starting ...

./job_3 sleeping for 5 seconds ...

./job_4 starting ...

./job_5 starting ...

./job_4 sleeping for 4 seconds ...

./job_5 sleeping for 2 seconds ...

./job_2 exiting ...

./job_6 starting ...

./job_6 sleeping for 2 seconds ...

./job_5 exiting ...

./job_7 starting ...

./job_7 sleeping for 1 seconds ...

./job_1 exiting ...

./job_8 starting ...

./job_8 sleeping for 3 seconds ...

./job_7 exiting ...

./job_9 starting ...

./job_9 sleeping for 5 seconds ...

./job_4 exiting ...

./job_6 exiting ...

./job_10 starting ...

./job_10 sleeping for 5 seconds ...

./job_3 exiting ...

./job_8 exiting ...

./job_9 exiting ...

./job_10 exiting ...

-(dearvoid@LinuxEden:Forum)-(~/tmp)-

[$$=6718 $?=0] ; bye