I don't understand the output from the following code:
public static void main(String[] args) {
int i1, i2, i3, i4;
byte b;
i1 = 128;
b = (byte) i1;
i2 = (int) b;
i3 = 0 | b;
i4 = 1 << 7;
System.out.format("i1: %d b: %d i2: %d i3: %d i4: %d\n", i1, b, i2, i3, i4);
}
Output:
i1: 128 b: -128 i2: -128 i3: -128 i4: 128
Because byte is an 8-bit two's-complement signed integer, the binary representations with a 1 in the most significant bit are interpreted as negative values, which is why b becomes -128, which I'm totally fine with. I also understand that it's probably a good idea to keep the interpretation consistent when casting, as with i2. But shouldn't i3 and i4 have identical bit patterns and therefore map to identical int values?
Sign extension is what is making i2 and i3 negative.
In the expression (0 | b), b is promoted to an int, and sign extension occurs as part of this promotion.
That's not happening in the expression assigned to i4. The constants 1 and 7 are already ints so there's no sign extension involved.
In this line:
i3 = 0 | b;
The "b" variable is automatically promoted to int type with sign extension because of the | operator, so becomes (int)-128, i.e. 0xffffff80.
When "or"ed with zero, its still the same value, namely -128
No, i4 is not a byte value, it's an int. That means that its sign bit is bit 31, not bit 7.
UPDATE: i3 is an int too, but it is initialized by extending a byte, so it keeps the sign from the byte value.
i2 = (int) b;
i3 = 0 | b;
the i3 statement is equivalent to:
i3 = 0 | ((int) b) = 0 | i2 so naturally it is going to have the same value as i2
That's simple. i3 = 0 | b; gets evaluated like byte, then it is converted to int. Whereas i4 = 1 << 7; will evaluate value as int and assign it to int. So in the first case we get 10000000b cast to int from byte, which will give us -128. And in the second we just assign this value to int without cast, which gives us 128.
If you want to get the unsigned value of the bit pattern in a byte:
b & 0xff
See also this old answer.
In
i3 = 0 | b;
I'm guessing the 0 | b part is evaluated as a byte, and the results are then cast to an int, while in
i4 = 1 << 7;
the 1 << 7 part is already an int.
The above guess has been pointed out in the comments to be wrong!
The correct version is: In the top expression, the b is already cast to an int with sign extension before the OR operation.
