Programming Tutorials

left shift operator, <<, in Java

By: Emiley J in Java Tutorials on 2007-09-07  

The left shift operator, <<, shifts all of the bits in a value to the left a specified number of times. It has this general form:
value << num

Here, num specifies the number of positions to left-shift the value in value. That is, the << moves all of the bits in the specified value to the left by the number of bit positions specified by num. For each shift left, the high-order bit is shifted out (and lost), and a zero is brought in on the right. This means that when a left shift is applied to an int operand, bits are lost once they are shifted past bit position 31. If the operand is a long, then bits are lost after bit position 63.

Java's automatic type promotions produce unexpected results when you are shifting byte and short values. As you know, byte and short values are promoted to int when an expression is evaluated. Furthermore, the result of such an expression is also an int.

This means that the outcome of a left shift on a byte or short value will be an int, and the bits shifted left will not be lost until they shift past bit position 31. Furthermore, a negative byte or short value will be sign-extended when it is promoted to int. Thus, the high-order bits will be filled with 1's. For these reasons, to perform a left shift on a byte or short implies that you must discard the high-order bytes of the int result. For example, if you left-shift a byte value, that value will first be promoted to int and then shifted. This means that you must discard the top three bytes of the result if what you want is the result of a shifted byte value. The easiest way to do this is to simply cast the result back into a byte.
The following program demonstrates this concept:

// Left shifting a byte value.
class ByteShift {
    public static void main(String args[]) {
        byte a = 64, b;
        int i;
        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

Since a is promoted to int for the purposes of evaluation, left-shifting the value 64 (0100 0000) twice results in i containing the value 256 (1 0000 0000). However, the value in b contains 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;

The program generates the following output:


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.

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Comments (2)


I did not understand the topic. Can u tell me what is the output for



And please explain. Thank you.


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