java - Why does 11010100 << 1 equal 110101000, not 10101000?

Question

Why when I try to shift bits for 11010100₂, the result is 110101000₂, not 10101000₂.

int a = Integer.parseInt("11010100", 2) << 1;

I try to do this:

int a = (byte)(Integer.parseInt("11010100", 2) << 1);

But if the output value is greater than 128, everything goes into minus, which is logical. How can I make that number of bits does not change?

question from:https://stackoverflow.com/questions/60321560/why-does-11010100-1-equal-110101000-not-10101000

Answer 1

Let's take it one step at a time.

1. Integer.parseInt("11010100", 2) - this is the int value 212. This is, by the way, needless; you can just write: 0b11010100.

2. 0b11010100 << 1 is the same as 0b110101000, and is 424.

3. You then cast it to a byte: (byte)(0b11010100 << 1). The bits beyond the first 8 all get lopped off, which leaves 0b10101000, which is -88. Minus, yes, because in java bytes are signed.

4. You then silently cast this -88 back up to int, as you assign it to an int value. It remains -88, which means all the top bits are all 1s.

Hence, the final value is -88.

If you want to see 168 instead (which is the exact same bits, but shown unsigned instead of signed), the usual trick is to use & 0xFF, which sets all bits except the first 8 to 0, thus guaranteeing a positive number:

byte b = (byte) (0b11010100 << 1);
System.out.println(b); // -88. It is not possible to print 168 when printing a byte.
int asUnsigned = b & 0xFF;
System.out.println(asUnsigned); // 168.

// or in one go:

System.out.println(((byte) (0b11010100 << 1)) & 0xFF); // 168