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Home/ Questions/Q 8821751
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T05:55:54+00:00 2026-06-14T05:55:54+00:00

I came across this problem in this website , and tried it in Eclipse

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I came across this problem in this website, and tried it in Eclipse but couldn’t understand how exactly they are evaluated.

    int x = 3, y = 7, z = 4;

    x += x++ * x++ * x++;  // gives x = 63
    System.out.println(x);

    y = y * y++;
    System.out.println(y); // gives y = 49

    z = z++ + z;
    System.out.println(z);  // gives z = 9

According to a comment in the website, x += x++ * x++ * x++ resolves to x = x+((x+2)*(x+1)*x) which turns out to be true. I think I am missing something about this operator precedence.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T05:55:55+00:00Added an answer on June 14, 2026 at 5:55 am

    Java evaluates expressions left to right & according to their precedence.

    int x = 3, y = 7, z = 4;
    
    x (3) += x++ (3) * x++ (4) * x++ (5);  // gives x = 63
    System.out.println(x);
    
    y = y (7) * y++ (7);
    System.out.println(y); // gives y = 49
    
    z = z++ (4) + z (5);
    System.out.println(z);  // gives z = 9
    

    Postfix increment operator only increments the variable after the variable is used/returned. All seems correct.

    This is pseudocode for the postfix increment operator:

    int x = 5;
    int temp = x;
    x += 1;
    return temp;
    

    From JLS 15.14.2 (reference):

    The value of the postfix increment expression is the value of the variable before the new value is stored.

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