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Home/ Questions/Q 7181555
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T17:34:55+00:00 2026-05-28T17:34:55+00:00

In Java, when we want to ensure that compiler should not do optimization by

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In Java, when we want to ensure that compiler should not do optimization by keeping a local copy of a variable, then we make the variable volatile. Using the variable as volatile ensures that the threads would not use a local copy of the variable but they would use the variable as it is stored in the main memory. But, does it mean that the volatile variable is thread-safe? Also how does it differ in case of a primitive type and in case we use a user defined object?

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  1. Editorial Team
    Editorial Team
    2026-05-28T17:34:56+00:00Added an answer on May 28, 2026 at 5:34 pm

    volatile means that the value will always be fresh; if another thread put a new object into the variable before you, you will see that object.

    It does not change the behavior of the value; you cannot magically make an object thread-safe.

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