ArithmeticException not thrown for overflow within 2^32

Question

I know assigning a number greater than 2^32 has a chance to generate an ArithmeticException but today while I was programming:

int x = 65535
System.out.println(x * x);

Output: -131071

So no exception but an unexpected result.

Answer 1

Overflow

Multiplication is not protected against overflows.

What you see here is integer overflow. If you take the biggest integer Integer.MAX_VALUE and add 1 you get the smallest integer INTEGER.MIN_VALUE:

int value = Integer.MAX_VALUE;
System.out.println(value);  // 2147483647
value++;
System.out.println(value);  // -2147483648

The same happens here because

65_535 * 65_535 = 4_294_836_225 > 2_147_483_647

---

Range of int

In Java int is a signed-32-bit value. In particular, it is not unsigned.

                 |    min-value   |   max-value   |
-----------------|----------------|---------------|
   signed-32-bit |          -2^31 |      2^31 - 1 |
                 | -2_147_483_648 | 2_147_483_647 |
-----------------|----------------|---------------|
 unsigned-32-bit |        2^0 - 1 |      2^32 - 1 |
                 |              0 | 4_294_967_295 |

---

Exception

A multiplication does not throw an ArithmeticException. To my knowledge this only happens if you divide by 0, since this should not be possible by definition. Also see the documentation of the exception.

For a protected multiplication consider using Math#multiplyExact (documentation).

Answer 2

I think you are confusing with how many bits the type int which is 32 bits to the numbers which it represent which is between:

 -2 147 483 648 <= int <= 2 147 483 647

Because it represents also negative numbers and there for can represent only 2^31

Answer 3

In java, int is primitive signed 32 bit which has

max value = 2.147.483.647

and

min value = -2.147.483.648

Result of your multiplication is 4.294.836.225.

Since u use int primitive type, If it overflows, it goes back to the minimum value and continues from there. If it underflows, it goes back to the maximum value and continues from there.

If u want to catch an exception, you can use Math class or Integer class.

        try
        {
          int c = Math.multiplyExact(a,b);
        } catch (ArithmeticException ex)
        {
            System.err.println("int is too small, falling back to long.");;
        }