Why conditional exit on error is executed without any error in this example code

Question

I am dealing with weird thing in my bash script and I can't understand what is happening here:

cd ~ && printf "CWD is $(echo $(pwd))\n" || printf "Critical err.\n" && exit 1;
printf "This should be printed, but it won't be.\n";

Bash 4.4.20.

But it should be executed only if the first part of pipe fails since there is "||". — TAKESHI KATAMANE, Apr 08 '20 at 09:15
You need to use bracket then try `(cd ~ && printf "CWD is $(echo $(pwd))\n") || (printf "Critical err.\n" && exit 1)` — Digvijay S, Apr 08 '20 at 09:19

score 0 · Accepted Answer · answered Apr 08 '20 at 09:21

0

It should be

(cd ~ && printf "CWD is $(echo $(pwd))\n") || (printf "Critical err.\n" && exit 1)

|| and && operands inside if condition (i.e. between round parentheses) are logical operands (or/and)

answered Apr 08 '20 at 09:21

Digvijay S

1 Answers1