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# shift Operators

Satish Kumar
Greenhorn

Joined: Aug 29, 2004
Posts: 3
This question is from khalid mughal page 84.. I could not solve this question.. pls help me with this..
<code>
public class Myclass
{
public static void main(String args[])
{
test(1<<32,"1<<32");
test(1<<31,"1<<31");
test(1<<30,"1<<30");
test(1,"1");
test(0,"0");
test(-1,"-1");
}
public static void test(int i,String exp)
{
if((i >> 1)!= (1>>>1))
System.out.pritln(exp);
}
}

a> "1<<32"
b> "1<<31"
c> "1"
d> "0"
e> "-1"
</code>
Vipin Das
Ranch Hand

Joined: Jul 05, 2004
Posts: 47
Hi anish,
a)If you shift an int x times the actual shifting occurs is x % 32.
Since 32 % 32 = 0, 1<<32 == 1<<0 == 1.
b)Another point to remember is >> and >>> behave same for a positive integer.

1. test(1<<32,"1<<32"); = test(1,"1<<32")// since 1<<32 = 1<<0 = 1

Inside the method i = 1;
i >> 1 and 1 >>>1 gets the same answer due to the point b above
2. test(1<<31,"1<<31")
Here the by shifting 32 times push the first bit to sign bit,makin it negative
thus it will look like 10000000........ To find out the real negative number
take the two's complement which will give -2147483648

inside the method i >> 1 = -1073741824 1>>> 1 =0
thus "1<<31" will be printed
3.test(1<<30,"1<<30");

this will 1<< 30 = 1073741824
inside the method i >>1 = 536870912. Again the condition becomes false and
the string "1<<30" will be printed.

4.test(1,"1"); This is same as case 1.
5.test(0,"0"); This makes the i >> 1 and 1 >>>1 the same
6.test(-1,"-1");
in this case the all the bits are set to one. ie; the bits in -1 = 11111111.....
Thus shifting it right using the signed >> to any number gives the same result -1 itself.
therefore inside test(), it the condition fails and the string "-1" will also be printed.

Thus there will be three answers,"1<<31", "1<<30", "-1"

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subject: shift Operators