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Home/ Questions/Q 7419561
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T08:05:34+00:00 2026-05-29T08:05:34+00:00

In java, with my knowledge, volatile variable make a thread reads/writes directly to main

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In java, with my knowledge, volatile variable make a thread reads/writes directly to main CPU (not in cache of each thread), so make its change visibles to other threads.

The thing I don’t know is : So, why this work (of volatile) can prevent compiler/CPU reorder statement of code.

thanks 🙂

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  1. Editorial Team
    Editorial Team
    2026-05-29T08:05:35+00:00Added an answer on May 29, 2026 at 8:05 am

    Here is a very good example illustrating the issue the prohibition on reordering is aimed to address (taken from here):

    class VolatileExample {
        int x = 0;
        volatile boolean v = false;
        public void writer() {
            x = 42;
            v = true;
        }
        public void reader() {
            if (v == true) {
                //uses x - guaranteed to see 42.
            }
        }
    }
    

    In this example, v is volatile, but x is not. If writer and reader are executed concurrently and the reader sees v set to true, x is guaranteed to be 42. Prior to Java-5, compiler was free to re-order the writes to x and v, so you could see x at zero after you’ve seen v set to true. This was confusing, and lead to subtle errors. Java-5 memory model addressed this issue by making volatile writes almost equivalent to synchronization.

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