bash read returns with exit code 1 even though it runs as expected

Question

I have this construct to initialize a variable with the contents of a file:

echo "yyy" > xxx
read -r -d '' PAYLOAD <<< $(cat xxx)
echo $?
echo $PAYLOAD

This results in:

1
yyy

Why is read returning 1? This is breaking my script, because I run with set -e.

Answer 1

read is returning 1 because it returns 0 only when end-of-file is not encountered.

As per help read:

Exit Status:
The return code is zero, unless end-of-file is encountered

You don't even need a read here, just use $(<file) to read the file content into a variable:

echo "yyy" > xxx
payload=$(<xxx)
echo $?
echo "$payload"

It is advisable to not use all uppercase variable names in order to avoid clash with ENV variables.

Answer 2

Actually the correct answer here is that in your case, read fails because it did not encounter a delimiter before EOF.

Because you set the delimiter to be the empty string (-d ''), read will fail to find that and will exit with 1. It will still read everything, but it only does this because it can't find the delimiter. That's why it appears to have worked, but seems to have failed also.

A trick proposed in another answer is to use a while loop/if statement. So something like this

if read -rd '' PAYLOAD; then :; fi <<< $(cat xxx)

Or (shorter)

if read -rd '' PAYLOAD; then :; fi < xxx

Should be enough to prevent the shell from seeing the exit code of the read command.

Sources: https://unix.stackexchange.com/a/265151/44793