How to capture and prefix to process output started within a bash script?

Question

I'm writing a small bash script where I am compiling 2 programs then executing them in the background within a bash script. These 2 programs do output some generic text. However I need to prefix these outputs like PROGRAM1: xxxxx. How do I achieve that? I have found several answers here however they weren't exactly applicable to this situation.

Here's the code:

#!/bin/bash
echo "This program compiles 2 programs, executes them (executes the 2nd one first, then sleeps for 0.01 second which then executes program 1), then observes the outputs"

gcc -O3 -std=c11 one.c -o one
gcc -O3 -std=c11 two.c -o two

./two &
sleep 0.01
TWO_PID=$(pgrep two)

./one $TWO_PID

#"Prefix output statements here"
#add code here

#Cleanup
rm -f one two

Diego Torres Milano · Accepted Answer · 2022-11-29 07:44:26Z

3

You can do something like this

#! /bin/bash

# ...

label() {
    while read -r l; do
        echo "$1: $l"
    done
}

./two | label "two" &
two_pid=$(pgrep two)
./one $two_pid | label "one"

You don't need the sleep and be careful as pgrep could match more than one process.

Also, instead of compiling and removing you should use make.

answered Nov 29, 2022 at 7:44

Diego Torres Milano

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3 Comments

cptalpdeniz Over a year ago

I need to do the sleep as it's requirement for a practice problem I'm working on.

cptalpdeniz Over a year ago

"Also, instead of compiling and removing you should use make." Do you mean to have a makefile in same directory and just call make within the bash script? Thanks for the answer!

Diego Torres Milano Over a year ago

Usually making and running are separate things but you can invoke make from the script if you want, at least it won't make it if sources didn't change.

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