On Sat, Dec 28, 2024 at 00:52:59 +0300, Павел Fortovey wrote:
> I apologize. I found out that this behavior is not a bug.
>
> I didn't think that bash is such a strange scripting language with its
> own interpretation of the concept of logic.
> Repeat-By:
>
> bash -c "true && true || false && echo bug"
An explanation is useful for the archives. In a command that chains
both the && and || operators, execution proceeds from left to right,
skipping or executing each internal command based on the value of $?
at that point, and the operator which precedes the command.
Taking the example:
true && true || false && echo bug
The first command, "true", is always executed. It sets $? to 0.
The operator after this command is && so we look at $? and execute
the next command only if $? is 0. Since the previous command did
set $? to 0, the second "true" is also executed. It sets $? to 0.
The next operator is || so we look at $? and only execute the next
command if $? is nonzero. Since $? is 0 at this point, we skip the
"false" command and proceed to the next operator.
The next operator is && so we look at $? and only execute the next
command if $? is zero. $? is 0 at this point (having been set that
way by the second "true" command), so we do in fact execute the
final command.
hobbit:~$ true && true || false && echo bug
bug
The final value of $? is whatever it was set to by the last command
executed. In my case, the echo successfully writes "bug" to stdout,
so $? is set to 0.
You might have been raised on some other language where the && and ||
operators have different precedence. In the shell language, they have
the same precedence.
In other words, it works like:
((true && true) || false) && echo bug
and not like:
(true && true) || (false && echo bug)
