How does Java handle the intermediate values while evaluating a long arithmetic expression?

I was wondering how does Java handle the intermediate values while evaluating a mathematical expression specially when the intermediate values exceeds their storage size. For an example,

int a = 1103515245;
int x = 1013904223;
int c = 2531011;

x = (a*x + c) & 0x7FFFFFFF; //  to make sure x will remain a positive integer.

Now, my question is while computing a*x, its value exceeds integer range (even during the addition it could happen), how this is taken care by Java Compiler?

Thank you.

>Solution :

JLS 15.17.1:

If an integer multiplication overflows, then the result is the low-order bits of the mathematical product as represented in some sufficiently large two’s-complement format.

JLS 15.18.2:

If an integer addition overflows, then the result is the low-order bits of the mathematical sum as represented in some sufficiently large two’s-complement format.

This is in contrast to C, which famously does not assume a two’s-complement architecture and treats overflow of signed types as undefined behavior.