Achieve Local Function

Question

What I want trying to achieve is to create a local function within a function. At the same time, the local function will not overwrite the outer function. Below is an example of a simple function and a nested function with argument to illustrate my problem.

#!/bin/bash usage() #<------------------------------- same function name {     echo "Overall Usage" }  function_A() {     usage() #<--------------------------- same function name     {         echo "function_A Usage"     }      for i in "$@"; do         case $i in             --help)                 usage                 shift                 ;;             *)                 echo "flag provided but not defined: ${i%%=*}"                 echo "See '$0 --help'."                 exit 0             ;;         esac     done }  function_A --help usage 

Here is output.

function_A Usage function_A Usage 

But what I want is

function_A Usage Overall Usage 

Is is possible to achieve without change their (functions) name and order? Please?

Note: I tried the local usage() but it seem not applicable to function.

Answer 1

Bash does not support local functions, but depending on your specific script and architecture you can control the scope of your function name through subshells.

By replacing the {..} with (..) in your definition, you'll get the output you want. The new definition of usage will be limited to the function, but so will e.g. any changes to variables:

#!/bin/bash usage()  {     echo "Overall Usage" }  function_A() (                  # <-- Use subshell     usage()     {         echo "function_A Usage"     }      for i in "$@"; do         case $i in             --help)                 usage                 shift                 ;;             *)                 echo "flag provided but not defined: ${i%%=*}"                 echo "See '$0 --help'."                 exit 0             ;;         esac     done )  function_A --help usage