Shell Script Exit Code - Unable to set

Question

I am running following test bash script:

test.sh

========

pass=$1
if [ $pass -eq 1 ]; then
   exit 0
else
   exit 1
fi

=============

So, If I run './test.sh 1', it should give me success code, i.e. 0. And if I run './test.sh 2' it should give me specific error code, i.e. 1.

But when I run the script, I am getting 0 as exit code for both the cases.

Output

========================

# ./test.sh 1 |echo $?
  0
# ./test.sh 2 |echo $?
  0
#

=========================

What am I doing wrong here? Any help will be greatly appreciated!

Noman A.

Answer 1

Your script works, your test is broken though. Don't use a pipe there.

# ./test.sh 1 ; echo $?
0
# ./test.sh 2 ; echo $?
1

What you proposed with a pipeline cannot work, because all the processes in pipeline are started "simultaneously". The shell starts a sub-shell to host each process (at least Bash does, implementations might vary -not sure about that), connects the input and output streams appropriately, then lets the OS schedule things as it sees fit.

So the rightmost process (in your case echo $?) is started at the "same time" as your test script. Therefore $? in that sub-shell (which will have been expanded before the actual process is started) can't possibly represent the return code from the test script - t.sh might not even have started yet!

See the Wikipedia article on Unix Pipelines for some more information, or your shells documentation on pipelines. (Bash Pipelines for instance.)