Consider the below java code.
Integer value = Integer.MIN_VALUE; System.out.println(value); value = -value; System.out.println(value); Output
-2147483648 -2147483648 How the negative value of Integer.MIN_VALUE value results the same value?
However the result can't be 2147483648 because the maximum value of Integer in java is 2147483647.
But want to know why -2147483648? What kind of bit-wise operations are happening internally?
When you negate -2147483648, it resolves to 2147483648, which exceeds the Integer.MAX_VALUE with 1. Then the value overflows to Integer.MIN_VALUE again.
From the JLS:
The integral types are byte, short, int, and long, whose values are 8-bit, 16-bit, 32-bit and 64-bit signed two's-complement integers.
So, every unary operation done on an integer will actually be applied on the two's complement representation of the number. When the Integer.MAX_VALUE is reached it will consist of a leading 0 and 31 1 bits. Adding 1 would make it a number with a leading 1 and 31 trailing 0s, which is actually the two's complement representation of Integer.MIN_VALUE.
When the Integer.MAX_VALUE is reached it will consist of 32 1 bits. Adding 1 would make it 33-bit number with a leading 1 and trailing 0s. - you got it a bit wrong. Integer.MAX_VALUE is a 0 followed by 31 1 bits. Adding one to it results in 1 followed by 31 0 bits, which is the twos complement representation of Integer.MIN_VALUEWhat kind of bit-wise operations are happening internally?
Java uses two's complement representation of signed numbers. Therefore, the change of sign operation, consists of two steps:
1 to the result.2147483648's representation is shown below:
10000000000000000000000000000000 Inverting it produces
01111111111111111111111111111111 Adding 1 makes it the same number again, i.e.
10000000000000000000000000000000 due to integer overflow.
1 is greater than what the highest positive value the int can hold, so that's an overflow.
Integer.MIN_VALUE. Duplicates aren't about the questions being the same, they are about This question already has an answer here, as the label states.