How can I execute a bash function using sudo?

Starting from the answer of bmargulies, I wrote a function to cover this issue, which basically realizes his idea.

# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #
# EXESUDO
# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #
#
# Purpose:
# -------------------------------------------------------------------- #
# Execute a function with sudo
#
# Params:
# -------------------------------------------------------------------- #
# $1:   string: name of the function to be executed with sudo
#
# Usage:
# -------------------------------------------------------------------- #
# exesudo "funcname" followed by any param
#
# -------------------------------------------------------------------- #
# Created 01 September 2012              Last Modified 02 September 2012

function exesudo ()
{
    ### ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ##
    #
    # LOCAL VARIABLES:
    #
    ### ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ##
    
    #
    # I use underscores to remember it's been passed
    local _funcname_="$1"
    
    local params=( "$@" )               ## array containing all params passed here
    local tmpfile="/dev/shm/$RANDOM"    ## temporary file
    local content                       ## content of the temporary file
    local regex                         ## regular expression
    
    
    ### ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ##
    #
    # MAIN CODE:
    #
    ### ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ##
    
    #
    # WORKING ON PARAMS:
    # ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    
    #
    # Shift the first param (which is the name of the function)
    unset params[0]              ## remove first element
    # params=( "${params[@]}" )     ## repack array
    
    
    #
    # WORKING ON THE TEMPORARY FILE:
    # ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    
    content="#!/bin/bash\n\n"
    
    #
    # Write the params array
    content="${content}params=(\n"
    
    regex="\s+"
    for param in "${params[@]}"
    do
        if [[ "$param" =~ $regex ]]
            then
                content="${content}\t\"${param}\"\n"
            else
                content="${content}\t${param}\n"
        fi
    done
    
    content="$content)\n"
    echo -e "$content" > "$tmpfile"
    
    #
    # Append the function source
    echo "#$( type "$_funcname_" )" >> "$tmpfile"
    
    #
    # Append the call to the function
    echo -e "\n$_funcname_ \"\${params[@]}\"\n" >> "$tmpfile"
    

    #
    # DONE: EXECUTE THE TEMPORARY FILE WITH SUDO
    # ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    sudo bash "$tmpfile"
    rm "$tmpfile"
}

Example of usage:
running the following snippet

#!/bin/bash

function exesudo ()
{
    # copy here the previous exesudo function !!!
}

test_it_out ()
{
    local params=( "$@" )
    echo "Hello "$( whoami )"!"
    echo "You passed the following params:"
    printf "%s\n" "${params[@]}" ## print array
}

echo "1. calling without sudo"
test_it_out "first" "second"

echo ""
echo "2. calling with sudo"
exesudo test_it_out -n "john done" -s "done"

exit

Will output

1. calling without sudo
   Hello yourname!
   You passed the following params:
   first
   second
   

2. calling with sudo
   Hello root!
   You passed the following params:
   -n
   john done
   -s
   foo
   

If you need to use this in a shell calling a function which is defined in your bashrc, as asked with a similar question on serverfault by another user, then you have to put the previous exesudo function on the same bashrc file as well, like the following:

function yourfunc ()
{
echo "Hello "$( whoami )"!"
}
export -f yourfunc

function exesudo ()
{
   # copy here
}
export -f exesudo

Then you have to logout and login again or use

source ~/.bashrc

Finally you can use exesudo as follow:

$ yourfunc
Hello yourname!

$ exesudo yourfunc
Hello root!

An alternative to calling your function with sudo is to just move the "sudo " calls inside your function. For example, I was wanting to set up a shortcut function in OS X for forwarding localhost to a particular port.

function portforward() {
    echo "y" | sudo ipfw flush;
    sudo ipfw add 100 fwd 127.0.0.1,$1 tcp from any to any 80 in;
    echo "Forwarding localhost to port $1!";
}

The function hits sudo and asks for my password. (Then pipes "y" to a prompt for ipfw, not related to this question). After that, sudo is cached so the rest of the function executes without the need to enter a password.

Essentially, this just runs like:

portforward 8000
Password:
Forwarding localhost to port 8000!

And fills my need because I only need to enter the password once and it's taken care of. Although it's a little ugly if you fail to enter the password the first time. Extra points for detecting whether the first sudo succeeded and exiting the function if not.

You can do that using declare -f, as in the following example:

function myfunc() {
    whoami
    echo First parameter is $1
}

myfunc foo
DECL=`declare -f myfunc`

sudo bash -c "$DECL; myfunc bar"

Each time you run sudo, it forks and execs a new copy of the shell, running as root. That shell does not inherit functions from your shell (it can't) and it doesn't inherit functions from previous executions. You will have to write out a file containing the function definition and invocation and sudo the invocation of that.