如何否定一个值与java [Integer.Min_Value]中的值相同

Question

How can these values be same in java?

-Integer.MIN_VALUE == Integer.MIN_VALUE

values are:

-2147483648 : -2147483648

i tried comparing it and returns true [Amazing!]

Answer 1

Yes, that's expected behaviour. The range of int is -2147483648 to +2147483647.

From the JLS section 15.15.4 (emphasis mine):

For integer values, negation is the same as subtraction from zero. The Java programming language uses two's-complement representation for integers, and the range of two's-complement values is not symmetric, so negation of the maximum negative int or long results in that same maximum negative number . Overflow occurs in this case, but no exception is thrown. For all integer values x , -x equals (~x)+1 .

~Integer.MIN_VALUE is Integer.MAX_VALUE ... and when you add one, it overlows to Integer.MIN_VALUE .

This is why when you implement a reversing comparator, you mustn't do this:

// BAD CODE!
public int compare(T x, T y) {
    return -originalComparator.compare(x, y);
}

Instead, use this:

// This is fine, assuming the comparator obeys its contract
public int compare(T x, T y) {
    return originalComparator.compare(y, x));
}

Answer 2

Look at the below code:

System.out.println(Integer.MIN_VALUE);
System.out.println(Integer.MAX_VALUE);
System.out.println(-Integer.MIN_VALUE);

The minimum value for integer is -2147483648 , and when you take a negation for this, it becomes 2147483648 , which is 1 greater than Integer.MAX_VALUE , so it goes out of range, and moves towards the other end, and become -2147483648

Answer 3

Java uses two's complement arithmetic, which means negating a value is the same as flipping its bits and adding 1.

But what about the number with bit pattern 100..00? When you flip its bits you get 011..11, and when you add 1 you get 100..00, the same number you started with!

For a range of other reasons this number is the lowest integer that can be represented, ie Integer.MIN_VALUE.