Java - PHP Number-Type Mix Up (9966006699)

I was doing some stuff with palindromes:

This number 9966006699 has been giving me problems. It's a product of 99979 and 99681

99979 * 99681 = 9966006699

I ran that in PHP

$i = 99979 * 99681;
echo $i;


9966006699 float(9966006699)

So in PHP the product is obviously a float data type. But in Java it's different as seen below :


 public static void main(String[] args) {

      float f = 99979 * 99681;

       long o = 99979 * 99681;

       double d = 99979 * 99681;

       int i = 99979 * 99681;




Google's calculator gives the right thing

I'm lost, why is Java giving the wrong output? and Does it have anything to do with the E9 stuff behind the float and double types? Help. Thanks


The numbers 99979 and 99681 are both int's. The multiplication expression is therefore an int expression too. The maximum value of an int expression is 2147...... Your value 9966006699 is way above that. Hence you have fallen in the realms of the strange behaviour that you get from modulo-n arithmetic. (That is, you have fallen victim to the C-family languages' version of the Y2K problem.)

Try this :

       long o = (long)99979 * 99681;

It looks like integer overflow as 9 966 006 699 > Integer.MAX_INT = 2 147 483 647. As int times int have type int the result overflows. Then it is cast to int/float/long etc.

This should be correct (for non-int):

long value = (long)99979 * (long)99681

Alternativly you can use BigInteger class which:

  1. May be slower the using int/long
  2. Don't have this problem for any numbers (long just moves the problem from 2^31 - 1 to 2^63-1).

Integer.MAX_VALUE is equal to 2^31-1, which is smaller than the number you're dealing with, so you're essentially getting integer overflow issues. You can get around this by using a long, or the BigInteger class.

You can read about the numeric limits of primitive data types here:

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