Instruction rearrangement under multithreading concurrency

In the case of multi-threaded concurrency , Instruction rearrangement will cause problems , Let's take a look at the following example ：

public class DemoCount {
    

    public static int r1, r2, x, y;

    public static void main(String[] args) throws InterruptedException {
    
        boolean f1, f2, f3, f4;
        f1 = f2 = f3 = f4 = false;
        int count = 0;
        while (true) {
    
            r1 = r2 = x = y = 0;
            Thread t1 = new Thread(() -> {
    
                x = 1;
                r1 = y;
            });

            Thread t2 = new Thread(() -> {
    
                y = 1;
                r2 = x;
            });

            t1.start();
            t2.start();
            t1.join();//  Ensure threads 0、1 Can execute main The thread prints the following output 
            t2.join();//  Ensure threads 0、1 Can execute main The thread prints the following output 

            ++count;

            if (r1 == 0 & r2 == 1 & !f1) {
    
                System.out.println(" The first " + count + " Occurrence results : r1=" + r1 + ":r2=" + r2);
                f1 = true;
            }
            if (r1 == 1 & r2 == 0 & !f2) {
    
                System.out.println(" The first " + count + " Occurrence results : r1=" + r1 + ":r2=" + r2);
                f2 = true;
            }
            if (r1 == 0 & r2 == 0 & !f3) {
    
                System.out.println(" The first " + count + " Occurrence results : r1=" + r1 + ":r2=" + r2);
                f3 = true;
            }
            if (r1 == 1 & r2 == 1 & !f4) {
    
                System.out.println(" The first " + count + " Occurrence results : r1=" + r1 + ":r2=" + r2);
                f4 = true;
            }
        }
    }
}

Running will produce four results ：

 The first 1 Occurrence results : r1=0:r2=1
 The first 59 Occurrence results : r1=1:r2=0
 The first 359000 Occurrence results : r1=1:r2=1
 The first 47518222 Occurrence results : r1=0:r2=0

Here's how , appear r1=0、r2=0 The case of indicates that there is an instruction reordering , We can check the code , The location of two code exchanges does not affect the result , So reordering can happen , For single threads , This reordering is no problem , But there will be problems in the case of multithreading concurrency , therefore r1=r2=0 It is because instruction reordering occurs in the case of multithreading .

From the above example, we can easily understand what is called Command rearrangement . To solve this problem, you can use primitives volatile Disable instruction reordering , The bottom layer is realized by four memory barrier instructions .