CODEWITHSUNDEEP
bashif-statementdemorgans-law
Edesem
Edesem

Reputation: 3

De Morgan's laws in Bash

I am writing a shell script. I have a getopts case statement initally and then I have another case statement that simply checks $@. In the second case statement I check if the user has typed --update or --uninstall.

In the first getopts case statement I have an error case:

"?") if  [[ $@ -ne "--uninstall" ]] && [[ $@ -ne "--update" ]]; then
     echo "Invalid option"

However it's not working as expected. I'm familiar with De Morgan's laws where !(A && B) = !A || !B and vice verrsa however this is not working in my script.

The expected behaviour would be I can run my script with the argument --uninstall or --update and run those functions. However not matter the configuration I have this if statement at best I can get one of the functions to run but not the other.

In the snippet I posted above running it with --uninstall it doesn't work, however --update works.

Upvotes: 0

Views: 115

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