Modifying the same variable using auto-increment/decrement more than once in the same statement gives undefined behaviour.

See Quantum Weirdness and the Increment Operator

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

Right out of perlop:

Note that just as in C, Perl doesn’t define when the variable is incremented or decremented. You just know it will be done sometime before or after the value is returned. This also means that modifying a variable twice in the same statement will lead to undefined behaviour. Avoid statements like: $i = $i ++; print ++ $i + $i ++;

An intellectual is someone whose mind watches itself.
-- Albert Camus

This is cool from an academic standpoint, but what valid piece of code actually uses this? To put it another way, what are you trying to do, apart from turn Perl into a pretzel?

Another question that comes to mind is, did you try the same thing, using brackets so as to guide the compiler to the order you want the operations performed?

How the result becomes as 8. If I execute this manually based on the operator precedence I suspect it should be 9. but I am wondering about the end result, which is 8. I hope, monks will have a better explanation.

Sorry for posting another "meta" reply, but... by any chance... are you doing (link @ GG) it again, jesuashok? Well, if not just don't take any offence, and the referenced links above may be of some interest to you.