Programs, Data, Variables, and Calculation - Beginning Java 2 SDK - page 60

The result of the expression is now zero because the third bit of indicators is zero.

To set a particular bit on, you can use the | operator, so to set the third bit in indicators on, you
can write:

indicators = indicators | 0x4; // Set the 3rd bit on

We can see how this applies to the last value we had for indicators:

Hexadecimal Binary

indicators 0xFF09 1111 1111 0000 1001

mask value 0x4 0000 0000 0000 0100

indicators | 0x4 0xFF0D 1111 1111 0000 1101

As you can see, the effect is just to switch the third bit of indicators on. All the others are unchanged. Of
course, if the bit were already on, it would stay on.

The bitwise operators can also be used in the op= form. Setting the third bit in the variable
indicators is usually written as:

indicators |= 0x4;

Although there is nothing wrong with the original statement we wrote, the one above is just a bit
more concise.

To set a bit off you need to use the & operator again, with a mask that has 0 for the bit you want as 0,
and 1 for all the others. To set the third bit of indicators off you could write:

indicators &= ~0x4; // Set the 3rd bit off

With indicators having the value 0xFF07 , this would work as follows:

Hexadecimal Binary

indicators 0xFF07 1111 1111 0000 0111

mask value 0x4 0000 0000 0000 0100

~0x4 0xFFFB 1111 1111 1111 1011

indicators & ~0x4 0xFF03 1111 1111 0000 0011

The ^ operator has the slightly surprising ability to interchange two values without moving either value
somewhere else. The need for this turns up most frequently in tricky examination questions. If you
execute the three statements: