In a Bash script, I'd like to define a boolean variable y to store the negated value of another boolean variable x. The following script,
#!/bin/bash
x=true
y=$(( ! "${x}" ))
echo "${y}"
sets variable y to 1. How can I change y so it evaluates to false instead?
Bash does not have the concept of boolean values for variables. The values of its variables are always strings. They can be handled as numbers in some contexts but that's all it can do.
You can use 1 and 0 instead or you can compare them (with = or ==) as strings with true and false and pretend they are boolean. The code will be more readable but they are still strings :-)
x=abc (without quotes) is not allowed but x=true is. Does Bash internally convert "boolean" values to their string representation always before storing them?
true and false are regular strings.
Commented
Dec 31, 2020 at 4:17
x=abc; echo $x should produce abc, and echo $? returns 0, indicating that that command executed successfully. What where you looking for? I'm only guessing here, but don't try to implement syntax you found handy in another language in your shell code. It will confuse future maintainers, etc. Good luck.
y=$(case "$x" in true ) echo false ;; * ) echo "Unknown Value for x=$x" 1>&2 ; echo "nonesuch" ;; esac)??true ; echo $?means? If you want to use return values0andnot 0, you'll need to translate from the number to the text that you want. Good luck.