Program starting other application

Mart777 · 04-04-2008, 01:09 PM

Hi,

My question is about writing a short program hat would be able to do the following:
- start other executable
- detect when that other executable completed its work

This short program would start other executable multiple times, e.g.: one time then wait till it finishes, modify output files and prepare new input files for that other executable then start again, and so on.

What would be good to use to write it to work under linux. I thought c++, is there any better language for this? Working on the files in between would be a bit extensive, so I thought c++ would be good for it. Do you know any place they explain how to do it to work under linux (we have CentOS 5).

Thank you.
-Mart

theNbomr · 04-04-2008, 01:46 PM

How about bash?

Code:

#! /bin/bash
while [ 1 ] ;
do
  echo starting
    ls -lash > dir.log
    sed -i s/^\\.// dir.log
    sleep 1
  echo ending
done;

It's what a shell is for.
--- rod.

beadyallen · 04-04-2008, 01:51 PM

Is there a reason you can't just use a shell script? Something like the following will work for simple cases:

Code:

#!/bin/sh

#This assumes you're program takes an inputfile called "input.dat", 
#and creates an ouput called "output.dat".
#Then you run progProcessResults to generate a new input file 
numruns=10
inputname="input.dat"
outputname="output.dat"

for ((i=1;i<=$numruns;i++))
do
  progmain $inputname $outputname
  progProcessResults $outputname $inputname
done

If you're sure you need to do it from within your own compiled program, have a look at 'man system'

edit: Beaten to it. Lol

Mart777 · 04-04-2008, 02:02 PM

Thank you!

I learned some basic shell script, but I see I will have to learn it more.
-Mart