Problem using functions with bash

isdigit · 12-24-2009, 03:42 PM

Code:

#!/bin/bash

#Functions
CallFunct() {
functio()
}

functio() {
echo -n
echo -n
}
#End 

# MAIN
CallFunct()

exit 0

When I try to run the current code, it stops on the bracket after functio() in CallFunct(). The only way I could get this to run was by removing CallFunct and just calling functio.
Could someone point me in the right direction?

evo2 · 12-24-2009, 03:49 PM

Hi,

what about this?

Code:

#!/bin/bash

#Functions
function CallFunct {
functio
}

function functio {
echo -n
echo -n
}
#End 

# MAIN
CallFunct

exit 0

isdigit · 12-24-2009, 03:54 PM

I don't know why it works now but here is the revised code:

Code:

#!/bin/bash

MainFunction() {
isRoot

} #Calling function; for simiplicity
isRoot() {
if [[ "$USER" != "root" ]]
 then
   echo "Sorry! You Have to be root to use this script"
   echo
   exit 1
fi
clear
} #Checks to see if user is titled root


MainFunction
exit  0

Edit: My kernel - 2.6.29.4 - Backtrack 4

GooseYArd · 12-24-2009, 07:21 PM

It works because (as evo2 pointed out, without saying) functio() wasn't declared when CallFunct() was declared. The order matters!