How to do a "if not" in Bash without a test?

Question

I would like to use the return value of a bash function in a 'if not' statement. The following is an example script:

#!/bin/bash

function myfunction () {
 if [ $1 = "one" ]; then
  return 1
 elif [ $1 = "two" ]; then
  return 2
 else
  return 0
 fi
}

if myfunction "two"; then
 # just using echo as an example here
 echo yep $?
else
 # just using echo as an example here
 echo nope $?
fi

Is there a way to modify the part 'if myfunction "two"' in a way that this script echoes 'yep 2'? I can only figure out this ugly way. How can can I solve this a better way?

Answer 1

if myfunction "two" in pseudocode is if the myfunction return code is zero when run with a single argument "two". If you want to reverse the comparison (is not zero) you can simply add a ! between if and myfunction.

Answer 2

Not quite sure what you're asking, but:

myfunction two; (( $? == 2 )) && echo yes || echo no

Answer 3

myfunction "two"
myvar=$?
if [ $myvar -gt 0 ]; then
 echo yep $myvar
else
 echo nope $myvar
fi

Answer 4

Why not do the echoing from within the function? This should be straightforward:

myfunction() {
    case "$1" in 
        one) echo "nope 0" ;;
        two) echo "yep 2" ;;
    esac
}
myfunction one
myfunction two