Can't access variable outside the function even it isn't local

Question

I have simple bash script. My bash version is GNU bash, version 4.4.20(1)-release (x86_64-redhat-linux-gnu)

#!/bin/bash
  
func() {
    nlvar="i'm not local variable"
    local lvar="i'm local variable"
}
echo $nlvar
echo $lvar

If I use

eval nlvar="i'm not local variable"

or

export nlvar="i'm not local variable"

still nothing is echoed

Is this the whole script? You defined the function. Where and how do you actually call it? — Kamil Maciorowski
– Kamil Maciorowski, Commented Dec 11, 2022 at 13:34
@KamilMaciorowski Yes, this is a whole script. I didn't call function. Just wanted to test the scope of variable. — kobayashi
– kobayashi, Commented Dec 11, 2022 at 15:40
If you don't call the function there is no scope except the global (undefined) one — Chris Davies
– Chris Davies, Commented Dec 11, 2022 at 15:51

Chris Davies · Accepted Answer · 2022-12-11 16:13:34Z

1

If you don't call the function there is no scope except the global (undefined) one:

#!/bin/bash
  
func() { :; }    # define a function

echo $nlvar
echo $lvar

Defining a function does not execute it, so at the point of execution of the echo statements neither variable has been set

answered Dec 11, 2022 at 16:13

Chris Davies

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