Exporting a variable from inside a function equals to global export of that variable?

Question

I use Ubuntu 16.04 with the native Bash on it.

I'm not sure if executing

#!/bin/bash
myFunc() {
    export myVar="myVal"
}
myFunc

equals in any sense, to just executing export myVar="myVal".

Of course, a global variable should usually be declared outside of a function (a matter of convention I assume, even if technically possible) but I do wonder about the more exotic cases where one writes some very general function and wants a variable inside it to still be available to everything, anywhere.

Would export of a variable inside a function, be identical to exporting it globally, directly in the CLI, making it available to everything in the shell (all subshells, and functions inside them)?

Solution

Your script creates an environment variable, myVar, in the environment of the script. The script, as it is currently presented, is functionally exactly equivalent to

#!/bin/bash

export myVar="myVal"

The fact that the export happens in the function body is not relevant to the scope of the environment variable (in this case). It will start to exist as soon as the function is called.

The variable will be available in the script's environment and in the environment of any other process started from the script after the function call.

The variable will not exist in the parent process' environment (the interactive shell that you run the script from), unless the script is sourced (with . or source) in which case the whole script will be executing in the interactive shell's environment (which is the purpose of "sourcing" a shell file).

Without the function call itself:

myFunc() {
    export myVar="myVal"
}

Sourcing this file would place myFunc in the environment of the calling shell. Calling the function would then create the environment variable.

See also the question What scopes can shell variables have?

Comments

[Noam Manos]

Note regarding subshells - If the function with the export myvar is called inside a ( subshell ), then the variable will still be unavailable in the parent scope.

[User Rebo]

What didn't work for me if i wrapped the export in (set -x export MY_VAR=...)

[Kusalananda]

@UserRebo It's unclear what you want to do. The exported variable will only be visible inside the subshell, and when the subshell terminates it will be gone. This is usually the effect one wants with an explicit subshell like this. Likewise, the effect of the set -x will also be limited the subshell.

[User Rebo]

Only wanted to make a note for other users. The set -x will prevent exporting variables in any scope. Didn't really refer to the subshell.

[Kusalananda]

@UserRebo set -x will absolutely not prevent exporting variables! It turns on tracing in the shell (i.e. the shell will produce debug output). It's the subshell that prevents the variable from being visible outside of the subshell.

[User Rebo]

I wrote a minimal script and variable didn't work when i added the set -x or added braces: export MYVAR=\first set -x export MYVAR=\second (export MYVAR=\third) echo "MYVAR: $MYVAR"

[Kusalananda]

@UserRebo Ah, now I see! You need ; (or a newline) between set -x and export. They are two separate commands. What you wrote originally, set -x export var=value would set $1 to the string export, and $2 to the string variable=value. I didn't spot the missing ; as it was such an odd command to give.

[User Rebo]

@NoamManos Nice thank you!