Nsryjdtyk

In my real project ,something like the code blew, read servel modules from file,like a1 a2 a3,a2 depends a1 and a3 depends a2.In another function like b1 b2 b3,but nothing to do with a1 a2 a3,so they can be parallel.

#!/bin/bash

dir=/root/

code_dir="code"

doubleit() 

{

    local transfer_dir=$1

    local moudles=`cat ${code_dir}/depend/$2`



    for module in $ moudles;do

        ${code_dir}/build/build_module.sh $transfer_dir $module

        transfer $transfer_dir

    done

}





export -f doubleit

export -f transfer



export code_dir



parallel doubleit ::: $dir ::: $A $B

If in function douleit,I called example.sh,do I need to export all the function in example.sh?

I got a wrong msg,can't figure out why.

Makefile:18: warning: overriding recipe for target `install'

common.mk:100: warning: ignoring old recipe for target `install'

upgrade_col.cpp: In function ‘void increase_update_count(unsigned int)’:

upgrade_col.cpp:170:33: warning: narrowing conversion of ‘uptime’ from ‘time_t {aka long int}’ to ‘unsigned int’ inside { } is ill-formed in C++11 [-Wnarrowing]

   update_ctx_t uctx = {1, uptime};

                                 ^

Makefile:18: warning: overriding recipe for target `install'

common.mk:100: warning: ignoring old recipe for target `install'

parallel: This job failed:

build_packages_parallel /root/v2.5/code/build/eps_cloud basis

when I run the jobs manually，echo $? just output 0,how does the parallel judge if the job is failed.I am very glad if there is any suggestion although it is closely related to my projects.

another question,if I run the function like this

doubleit arg1 &

doubleit arg2 &

wait

does this use the differnt cpu cores to speed up?
if one job failed,how to exit the shell function and killed all the bg task,I know $!,and wait the array,but not work.