Precedence of operators in Java -

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this question has answer here:

when run java code:

int[] = new int[10];     int = 0,j = 0;     while(i < 10){         a[i++] = j+++j++;     }     system.out.println(arrays.tostring(a));

i output: [1, 5, 9, 13, 17, 21, 25, 29, 33, 37]. can please explain how statement a[i++] = j+++j++ resolved.

the first j++ in expression j+++j++ increments j , returns previous value.

the second j++ increments j , returns previous value, value after first j++ incremented it.

at start of next iteration, j larger 2 compared value @ start of previous iteration (since previous iteration incremented j twice).

therefore :

a[0] = 0++ + 1++ = 0 + 1 = 1; a[1] = 2++ + 3++ = 2 + 3 = 5; a[2] = 4++ + 5++ = 4 + 5 = 9; ...

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