Argthtjtr

Recently I found out that if you take a double or float and divide it by 0 that you get infinity as value. I converted infinity to every datatype (by writing (byte)variable, (short)variable etc.) and I found out that if you convert it into byte or short it gives the value -1, but if you convert it into any other datatype it gives you the max value of it back. I was wondering why that happened. Anyone got a guess?

Here is the code im using to test it.

double dsa = 1.00 / 0;



System.out.println("Byte");

System.out.println((byte)dsa);

System.out.println(Byte.MAX_VALUE + "n");



System.out.println("Short");

System.out.println((short)dsa);

System.out.println(Short.MAX_VALUE + "n");



System.out.println("Integer");

System.out.println((int)dsa);

System.out.println(Integer.MAX_VALUE + "n");



System.out.println("Long");

System.out.println((long)dsa);

System.out.println(Long.MAX_VALUE + "n");



System.out.println("Float");

System.out.println((float)dsa);

System.out.println(Float.MAX_VALUE + "n");



System.out.println("Double");

System.out.println((double)dsa);

System.out.println(Double.MAX_VALUE + "n");

And here is what the console gives me:

Byte

-1

127



Short

-1

32767



Integer

2147483647

2147483647



Long

9223372036854775807

9223372036854775807



Float

Infinity

3.4028235E38



Double

Infinity

1.7976931348623157E308

Because positive infinity is defined as 0x7ff0000000000000. From Double class source code:

/**

 * A constant holding the positive infinity of type

 * {@code double}. It is equal to the value returned by

 * {@code Double.longBitsToDouble(0x7ff0000000000000L)}.

 */

public static final double POSITIVE_INFINITY = 1.0 / 0.0;

Also please note that casting is not converting.