vikkyhacks - 1 year ago 151
Java Question

# How is 0x80000000 equated to -2147483648 in java?

Taking the binary of

`0x80000000`
we get

``````1000 0000 0000 0000 0000 0000 0000 0000
``````

How does this equate to
`-2147483648`
. I got this question with this program.

``````class a
{
public static void main(String[] args)
{
int a = 0x80000000;
System.out.printf("%x %d\n",a,a);
}
}

meow@VikkyHacks:~/Arena/java\$ java a
80000000 -2147483648
``````

EDIT I learned that 2's complement is used to represent negative numbers. When I try to equate this with that 1's complement would be

``````1's Comp. :: 0111 1111 1111 1111 1111 1111 1111 1111
2's Comp. :: 1000 0000 0000 0000 0000 0000 0000 0000
``````

which again does not make any sense, How does
`0x80000000`
equate to
`-2147483648`

This is what happens with signed integer overflow, basically.

It's simpler to take `byte` as an example. A `byte` value is always in the range -128 to 127 (inclusive). So if you have a value of `127` (which is 0x7f) if you add 1, you get -128. That's also what you get if you cast 128 (0x80) to `byte`:

``````int x = 0x80; // 128
byte y = (byte) x; // -127
``````

Overflow (in 2s complement integer representations) always goes from the highest expressible number to the lowest one.

For unsigned types, the highest value overflows to 0 (which is again the lowest expressible number). This is harder to show in Java as the only unsigned type is `char`:

``````char x = (char) 0xffff;
x++;
System.out.println((int) x); // 0
``````