I have that sample script:
#!/bin/sh
while read ll </dev/fd/4; do
echo "1 "$ll
while read line; do
echo $line
read input </dev/fd/3
echo "$input"
done 3<&0 <notify-finished
done 4<output_file
Currently The first loop do not iterate just stays on line 1. How do I fix that without bashisms because it has to be highly portable. Thanks.
Your code already has bashisms. Here, I'm taking them out (and simplifying the FD handling for better readability):
#!/bin/sh
while read ll <&4; do # read from output_file
printf '%s\n' "1 $ll"
while read line <&3; do # read from notify-finished
printf '%s\n' "$line"
read input # read from stdin
printf '%s\n' "$input"
done 3<notify-finished
done 4<output_file
Run the script as follows:
echo "output_file" >output_file
echo "notify-finished" >notify-finished
echo "stdout" | ./yourscript
...and it correctly exits with the following output:
1 output_file
notify-finished
stdout
Notes:
echo's behavior is wildly nonportable across POSIX platforms. See the APPLICATION USAGE section of the POSIX spec for echo, which advises using printf instead./dev/fd/## is not specified by POSIX; it is an extension made available both by Linux distributions (creating a symlink to /proc/self/fd -- /proc being itself an unspecified extension) and by bash itself. Use <&4 in place of </dev/fd/4.-r argument to read -- which is POSIX-specified, and prevents the default behavior of treating backslashes as escape sequences for newlines and characters in IFS. Without it, foo\bar is read as foobar, thus not reading your data as it truly exists in its input sources.
/dev/fdis either a Linuxism or a bashism (bash emulates it on non-Linux operating systems, but not all shells do), so you're already nonportable. Use<&4, not</dev/fd/4.output_fileandnotify-finished, so it will run in a way that successfully demonstrates your bug without the person who wants to see your problem needing to do anything else.</dev/fd/4notation that was causing the trouble. If you second that comment as an answer, I'll accept it. Thank you.