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
Double equals ( == ) will perform a type conversion when comparing two things, and will handle NaN , -0 , and +0 specially to conform to IEEE 754 (so NaN != NaN , and -0 == +0 );
If you just check if two floats subtracted from each other are less than E, but the difference comes out at -1, it will count it as being equal. We use the absolute value to ensure we can only get a value less than epsilon if the result is close to 0. Depending on your usage, the value of epsilon can vary.
Using the == Operator As a result, we can't have an exact representation of most double values in our computers. They must be rounded to be saved. In that case, comparing both values with the == operator would produce a wrong result.
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; // -128
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
This is the case when there is overflow, with respect to the range of a data type. Here is an example that I can share.
int number = 0x80; // 0x80 is hexadecimal for 128 (decimal)
byte castedNumber = (byte)(number); // On casting, there is overflow,as byte ranges from -128 to 127 (inclusive).
System.out.println(castedNumber); //Output is -128.
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