shell里实现‘多线程’

分类:shell编程 阅读:5970 次

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:

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

-(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