calling execu in a bash while loop, not working

Question

I have a bash script that has to run the same file (c execut) but with different argument of input and save the output files of the c simulation in different folders. I wrote this but I don't know why it is not working!

mu=1
num=0
while test $num -le 6;do
    mkdir $num
    cd $num
    while test $mu -le 100; do
            ./infosimul "$num" "$mu" 
            mu=$((mu+3))
        done
    mu=0
    cd ../
    num=$((num+1))
done

and the oputput is:

testBash.sh: 7: testBash.sh: ./infosimul: not found

but if I run just

./infosimul: 1 3

works properly.

I am an UBUNTU user!

You change the dir by doing cd $num So infosimul is not there in the current directory. — P.P
– P.P, Commented May 20, 2014 at 10:56

dogbane · Accepted Answer · 2014-05-20 10:55:28Z

3

Your script is changing directory by calling cd $num and the infosimul program doesn't exist in the new directory.

Try calling ../infosimul instead or specify the full path to infosimul.

answered May 20, 2014 at 10:55

dogbane

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