Question

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?

Answer 1

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.