Running Executable in Bash Script
 

Hey guys, so I've been trying to write a bash script called runSorter.sh that runs an executable that also takes in some parameters and outputs the results to a text file. The executable, sorter, takes in a number parameter. I want to make it so that you can input as many number parameters into runSorter.sh as you want and it will run the sorter executable for each one. So far, what I have looks like this:

#!/bin/bash
args=("$@")
INDEX=0

if [ -z args ]; then
echo "Error"

else
while [ $# -gt $INDEX ]; do
NUM=${args[$INDEX]}
echo $NUM
echo ./sorter $NUM
let INDEX=INDEX+1
done
fi


My problem is that when I run ./run-sorter.sh 100 on my terminal, it just prints this to the screen:
./sorter 100

How can I have so that it properly executes sorter and outputs everything to a text file? Thanks in advance.

Change this:
to

Hmmm, that works but now it prints sorter's results on another line. How do I get it to print the results on the same line as echo $NUM?

Change `echo $NUM` to `echo -n $NUM`.

Oooh, I see. Thank you very much.