Java Integer Max Range multiplication

Question

Integer num = 2147483647;
Integer res =  num * num;
System.out.println(res);

The Out put for above is 1. Am not sure why. Can someone please explain.

Thanks in advance.

Integers have a limited range of 2^32-1... Roughly 2 billion — Trey Shaffer, Mar 05 '15 at 06:09
2^32-1 for unsigned integers, (2^32-1)/2 for signed integers — Trey Shaffer, Mar 05 '15 at 06:10
Let me re-phrase my Question. I know it goes out of bounds for Integer. But Why result is one. What logic ? I had a look at Integer.java class, but I didn't find anything useful. — Aditya, Mar 05 '15 at 08:02

score 1 · Answer 1 · answered Mar 05 '15 at 06:13

1

This is supposed to demonstrate why result = 1:

    long x = Integer.MAX_VALUE;
    long y = Integer.MAX_VALUE;
    long res = x * y;
    System.out.println(Long.toHexString(res));

prints

3fffffff00000001

if we cast res to int we will get 1

answered Mar 05 '15 at 06:13

Evgeniy Dorofeev

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score 0 · Answer 2 · answered Mar 05 '15 at 06:10

0

It's because of Integer overflow. Java has a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647 (inclusive) and your result(res) is beyond Integer max range.

answered Mar 05 '15 at 06:10

Arpit Aggarwal

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score 0 · Answer 3 · edited May 23 '17 at 10:09

0

this because of the flag value
0: iconst_0
1: istore_1

As the limit is out of range for Integer it will set the flag of 1 for more about how it works use this link

edited May 23 '17 at 10:09

Community

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answered Mar 05 '15 at 06:10

Bhargav Modi

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score 0 · Answer 4 · answered Mar 05 '15 at 06:11

That's because it overflows the range of integer and even long which is between -2,147,483,648 and a maximum value of 2,147,483,647 (inclusive).

You should try using BigInteger if that's the case like:

String num = "2147483647";
BigInteger mult = new BigInteger(num);
System.out.println(mult.multiply(mult));

Java Integer Max Range multiplication

4 Answers4