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.

Answer 2

$(...) captures the text sent to standard output by the command contained within. return does not output to standard output. $? contains the result code of the last command.

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