Search code examples
javajvmbytecodesynchronizedspecifications

Why do we need to call 'monitorexit' instruction twice when we use 'synchronized' keyword?

According to JVM Specification, paragraph 3.14 Synchronization, JVM call monitorexit instruction twice.

I wonder why JVM need such behavior? Why do we call this instruction 2 times, but not 3 or 4 or N times?

May be it's somehow connected with types of synchronization locks?

Solution

  • The code is not "calling" the monitorexit instruction twice. It is executing it once on two different code paths.

    • The first codepath is for when the code in the synchronized block exits normally.
    • The second codepath is in an implicit exception handling path for the case where the block terminates abnormally.

    You could write the bytecodes in the example as pseudo-code that looks like this:

    void onlyMe(Foo f) {
    monitorEntry(f);

    try {
        doSomething();
        monitorExit();
    } catch (Throwable any) {
        monitorExit();
        throw any;
    }
}

    If you want more information on this, take a look at this older question: