Java The Complete Reference, Seventh Edition

i = a << 2;

b = (byte) (a << 2);

System.out.println("Original value of a: " + a);

System.out.println("i and b: " + i + " " + b);

}

}

The output generated by this program is shown here:

Original value of a: 64

i and b: 256 0

Sinceais promoted tointfor the purposes of evaluation, left-shifting the value 64

(0100 0000) twice results inicontaining the value 256 (1 0000 0000). However, the value

inbcontains 0 because after the shift, the low-order byte is now zero. Its only 1 bit has

been shifted out.

Since each left shift has the effect of doubling the original value, programmers frequently

use this fact as an efficient alternative to multiplying by 2. But you need to watch out. If you

shift a 1 bit into the high-order position (bit 31 or 63), the value will become negative. The

following program illustrates this point:

// Left shifting as a quick way to multiply by 2.

class MultByTwo {

public static void main(String args[]) {

int i;

int num = 0xFFFFFFE;

for(i=0; i<4; i++) {

num = num << 1;

System.out.println(num);

}

}

}

The program generates the following output:

536870908

1073741816

2147483632

-32

The starting value was carefully chosen so that after being shifted left 4 bit positions, it

would produce –32. As you can see, when a 1 bit is shifted into bit 31, the number is interpreted

as negative.

The Right Shift

The right shift operator,>>, shifts all of the bits in a value to the right a specified number of

times. Its general form is shown here:

value >> num

66 Part I: The Java Language