[SOLVED] echo to stdout if a sub-function succeeds, suppress stderr if it fails

Issue

Inside of a bash script, I want to define a function to do the following:

1. invoke a sub-program (/usr/sbin/i2sget)
2. if the sub-program succeeds, echo what it printed to stdout
3. if the sub-program fails, suppress stderr and print "SUPPRESSED" on stdout

Here's what I have that doesn't quite work. (Note: i2sget returns a '0' on success and '2' on a read error):

read_i2s_register () {
    local RESULT=`/usr/bin/i2sget -y 1 0x55 0x08`
    if [ $? -eq 0 ]: then
        echo ${RESULT}
    else
        echo "SUPPRESSED"
    fi
}

When /usr/bini2sget succeeds, the above function properly echoes its stdout (captured in ${RESULT}.

But when it fails, it prints "Error: Read failed" (presumably on stderr) and it doesn't print SUPPRESSED on stdout.

What's the right way to accomplish the desired behavior?

Solution

The exit code check won't work when you add local because that has its own exit status that masks whatever the command exited with. To address this, you have to separate the two pieces:

local RESULT
# change 2>/dev/null to 2>&1 if you want
# stderr captured in the RESULT
RESULT=$( /usr/bin/i2sget -y 1 0x55 0x08 2>/dev/null ) 
if [[ $? -eq 0 ]] ; then
  echo "$RESULT"
else
  echo SUPPRESSED
fi

Or a little more succinctly to avoid an antipattern:

local RESULT
# change 2>/dev/null to 2>&1 if you want
# stderr captured in the RESULT
if RESULT=$( /usr/bin/i2sget -y 1 0x55 0x08 2>/dev/null ) ; then
  echo "$RESULT"
else
  echo SUPPRESSED
fi

Answered By - tjm3772

Answer Checked By - Senaida (WPSolving Volunteer)