Shell中实现“多线程”执行脚本文件完美解决方案

job_1
job_2
job_2
.....
job_100

-(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