How can I assign the output of a function to a variable using bash?


I have a bash function that produces some output:

function scan {
  echo "output"

How can I assign this output to a variable?

ie. VAR=scan (of course this doesn't work - it makes VAR equal the string "scan")

11/27/2009 5:36:08 PM

Accepted Answer


Exactly the same way as for programs.

11/27/2009 5:37:57 PM

You may use bash functions in commands/pipelines as you would otherwise use regular programs. The functions are also available to subshells and transitively, Command Substitution:


Is the straighforward way to achieve the result you want in most cases. I will outline special cases below.

Preserving trailing Newlines:

One of the (usually helpful) side effects of Command Substitution is that it will strip any number of trailing newlines. If one wishes to preserve trailing newlines, one can append a dummy character to output of the subshell, and subsequently strip it with parameter expansion.

function scan2 () {
    local nl=$'\x0a';  # that's just \n
    echo "output${nl}${nl}" # 2 in the string + 1 by echo

# append a character to the total output.
# and strip it with %% parameter expansion.
VAR=$(scan2; echo "x"); VAR="${VAR%%x}"

echo "${VAR}---"

prints (3 newlines kept):



Use an output parameter: avoiding the subshell (and preserving newlines)

If what the function tries to achieve is to "return" a string into a variable , with bash v4.3 and up, one can use what's called a nameref. Namerefs allows a function to take the name of one or more variables output parameters. You can assign things to a nameref variable, and it is as if you changed the variable it 'points to/references'.

function scan3() {
    local -n outvar=$1    # -n makes it a nameref.
    local nl=$'\x0a'
    outvar="output${nl}${nl}"  # two total. quotes preserve newlines

VAR="some prior value which will get overwritten"

# you pass the name of the variable. VAR will be modified.
scan3 VAR

# newlines are also preserved.
echo "${VAR}==="




This form has a few advantages. Namely, it allows your function to modify the environment of the caller without using global variables everywhere.

Note: using namerefs can improve the performance of your program greatly if your functions rely heavily on bash builtins, because it avoids the creation of a subshell that is thrown away just after. This generally makes more sense for small functions reused often, e.g. functions ending in echo "$returnstring"

This is relevant.

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow