How to exit a function in bash

Question

How would you exit out of a function if a condition is true without killing the whole script, just return back to before you called the function.

Example

Answer 1

If you want to return from an outer function with an error without exiting you can use this trick:

do-something-complex() {
  # Using `return` here would only return from `fail`, not from `do-something-complex`.
  # Using `exit` would close the entire shell.
  # So we (ab)use a different feature. :)
  fail() { : "${__fail_fast:?$1}"; }

  nested-func() {
      try-this || fail "This didn't work"
      try-that || fail "That didn't work"
  }
  nested-func
}

Trying it out:

$ do-something-complex
try-this: command not found
bash: __fail_fast: This didn't work

This has the added benefit/drawback that you can optionally turn off this feature: __fail_fast=x do-something-complex.

Note that this causes the outermost function to return 1.

Answer 2

Use return operator:

function FUNCT {
  if [ blah is false ]; then
    return 1 # or return 0, or even you can omit the argument.
  else
    keep running the function
  fi
}

Answer 3

Use:

return [n]

From help return

return: return [n]

Return from a shell function.

Causes a function or sourced script to exit with the return value
specified by N.  If N is omitted, the return status is that of the
last command executed within the function or script.

Exit Status:
Returns N, or failure if the shell is not executing a function or script.