Thread: Compiler reorder the statement to excute

I'm reading an ebook. they're using volatile keywords for a variable.
This is my example code:

Code:

volatile boolean check;
int a ;
//other function
public void functionA(){
     check = true;
     while(a>0){ a--;}
}

They explain check variable is volatile because: the compiler might decide to reorder the statements in function A if it recognizes that there are no dependencies. It is allowed to do this if it thinks it will make the code execute faster
It means in real, this code might run like:

Code:

while(a>0){ a--;}
check = true;

Does this statement true ? I feel very strange of this because I have never heard this before.

thanks

Yes, optimisation is allowed as long as the compiler can determine that the result will produce observable behaviour that is the same as the observable behaviour without the optimisation.

However, the ebook is misinforming you. The volatile keyword is insufficient to prevent the compiler performing those optimisations. There is no guarantee that check must be set true before a is changed.

It could also change the code to

Code:

    check = true;
    a = 0;

or to

Code:

    a = 0;
    check = true;

It can't quite optimise that loop to a=0; because a might start out negative. However something simewhat similar that accounts for that could be used, assuming it hasn't proved that a is never negative.

But yeah there be dragons ahead. I'm not sure you want to get mixed up in this stuff. Take the blue pill.

Edit - duh, never mind. A decent compiler will conditionally set a to 0 if it's greater than 0 and leave it alone for negative values. No need to run the loop, though.