Prasanna Kumar Prasanna Kumar - 1 year ago 64
Java Question

Why maximum integer number multiplication gives 1 as result

I think much explanation is not required, why below calculation gives result as 1?

int a = 2147483647;
int b = 2147483647;

int c = a * b;
long d = a * b;
double e = a * b;

System.out.println(c); //1
System.out.println(d); //1
System.out.println(e); //1.0

Answer Source

The binary representation of the integer number 2147483647 is as following:

01111111 11111111 11111111 11111111

Multiplying this with itself results in the number 4611686014132420609 whose binary representation is:

00111111 11111111 11111111 11111111 00000000 00000000 00000000 00000001

This is too large for the int type which has only 32 bits. The multiplication of a * b is done as an integer multiplication only, regardless of the variable's type to which the result is assigned (which might do a widening conversion, but only after the multiplication).

So, the result simply cuts off all bits that do not fit into the 32 bits, leaving only the following result:

00000000 00000000 00000000 00000001

And this simply is the value 1.

If you want keep the information, you must do the multiplication with the long type which has 64 bits:

long a = 2147483647;
long b = 2147483647;
long mult = a * b;

System.out.println((int) mult);     // 1
System.out.println(mult);           // 4611686014132420609
System.out.println((double) mult);  // 4.6116860141324206E18

If you need more bits for a calculation you might consider BigInteger (for integral numbers) or BigDecimal (for decmial numbers).

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