Understanding Bash short-circuiting

Question

First of all I'm not a Bash pro. I discovered few months ago that if I use both the && and || short circuit operators in sequence with curly braces, then in case the first statement exits with a truthful value, if the last statement in the true block exits non-zero, then the fail block will be executed too. Like this:

returnNumber 0 && {
    echo 'OK'
    returnNumber 1
} || {
    echo 'NG'
}

Will output:

OK
NG

So, I looked for the easiest solution for this, and came up with this:

returnNumber 0 && {
    echo 'OK'
    returnNumber 1
    :
} || {
    echo 'NG'
}

I know, it is easy to leave out the colon builtin, but is it a proper way for a workaround?

Answer 1

This is actually a very common Bash pitfall. It is not a bug.

returnNumber 0 evaluates to true, so the second block (joined by logical and &&) is evaluated as well to make sure the result of first && second is still true.
The second block outputs OK but evaluates to false, so now the result of first && second is false. This means that the third portion (joined by logical or ||) must be evaluated as well, causing NG to be displayed as well.

---

Instead of relying on && and ||, you should be using if statements:

if returnNumber 0; then
    echo 'OK'
    returnNumber 1
else
    echo 'NG'
fi

tl;dr: Never use x && y || z when y can return a non-zero exit status.