variable within functions-Linux Shell Script

Question

i am facing an issue with this simple code , where i call a function , but the variable $X of that function doesnt seem to pass outside, since the printout message is always "Your interface is ." ... Am i missing something?

#! /bin/bash
function choose
{
echo -e " Choose your interface:"
echo -e " 1) WLan0"
echo -e " 2) WLan0mon"
echo -e " Choose: "
read -e X
if [ "$X" = "1" ]
then
    X="wlan0"
elif [ "$X" = "2" ]
then
    X="wlan0mon"
fi
}


(choose)
echo -e "Your interface is $X."

I think X only exists inside the scope of your choose function. Try either removing the function definition/call altogether, or moving the echo inside the body of the choose function. — Mark
– Mark, Commented Apr 16, 2017 at 17:28

Phil Grigsby · Accepted Answer · 2017-04-16 14:55:45Z

2

Running (choose) using the parentheses executes it in a subshell.

From Advanced Bash-Scripting Guide:

Variables in a subshell are not visible outside the block of code in the subshell. They are not accessible to the parent process, to the shell that launched the subshell. These are, in effect, local variables.

Removing the parentheses around choose will make variable X visible to your echo.

answered Apr 16, 2017 at 14:55

Phil Grigsby

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