List contains returns true for checking if 2 exists in a list inserted with "2"

[1sand0s]

Minimum working example

#!/bin/bash                                                                                         
. src/util/list.sh

lst=$(List)
$lst add 1
$lst add 3
$lst add "2"
$lst contains 2
echo $? # returns 0

This line essentially always does string compare which is true even when comparing "2" with 2.

if [ "${el}" = "${val}" ]; then 
	return $TRUE
fi

[gliga]

Thanks for the comment.

That example looks like something I would expect to see in bash. Everything is string in a way; there is really no difference between "2" and 2 when used in bash line, like you did.

$lst add "2"

is no diff from

$lst add 2

Here is also more "native" bash example.

#!/bin/bash                                                                                                                                                                                   

x=2
z="2"

if [ ${x} = "${z}" ]; then
        echo "same"
fi

What do you expect to see as the result in this case?

[gliga]

Closing based on the comment above and in-person agreement.