bash: How can I detect if the previous command failed, then exit - from inside a function?

Question

I am trying to write a function in a bash script which accepts a command as an argument, executes the command -- AND -- if the command non-zero exited, abort the script:

file: command_wrapper.sh

#!/usr/bin/env bash

function exec_cmd() {
    local command=${1}
    $command || exit            # run the command if $? is 1 exit
    if [[ $? -eq 1 ]]           # AGAIN if $? is 1 exit
    then
        echo "command failed"
        exit
    fi

    return 0
}

exec_cmd "grep -R somsomeoemoem ."      # call the function, passing a 
                                        # command that will exit with 1

echo "rest of script running"

call it:

% ./command_wrapper.sh
./command_wrapper.sh:exec_cmd "grep -R somethingyouwillneverfind ."
rest of script running

the script carries on after the function is run with a command that non-zero exits - why?
How can I get this command to error exit when $command fails?

Answer 1

Your function works as designed but the issue is your assumption is incorrect. grep -R somsomeoemoem . doesn't exit in error.

Try with this command to see the function exiting:

exec_cmd "false"