Return value in a Bash function

Question

I am working with a bash script and I want to execute a function to print a return value:

function fun1(){
  return 34
}
function fun2(){
  local res=$(fun1)
  echo $res
}

When I execute fun2, it does not print "34". Why is this the case?

Answer 1

Although bash has a return statement, the only thing you can specify with it is the function's own exit status (a value between 0 and 255, 0 meaning "success"). So return is not what you want.

You might want to convert your return statement to an echo statement - that way your function output could be captured using $() braces, which seems to be exactly what you want.

Here is an example:

function fun1(){
  echo 34
}

function fun2(){
  local res=$(fun1)
  echo $res
}

Another way to get the return value (if you just want to return an integer 0-255) is $?.

function fun1(){
  return 34
}

function fun2(){
  fun1
  local res=$?
  echo $res
}

Also, note that you can use the return value to use boolean logic like fun1 || fun2 will only run fun2 if fun1 returns a non-0 value. The default return value is the exit value of the last statement executed within the function.