shift question

[bani kaali]

The following code will print
1: int i = 1;
2: i <<= 31;
3: i >>= 31;
4: i >>= 1;
5:
6: int j = 1;
7: j >>= 31;
8: j >>= 31;
9:
10: System.out.println("i = " +i );
11: System.out.println("j = " +j);
Ans is i = -1 and j =0;
could any one tell me how i is -1 ?

[Chung Huang]

modify your code as follows:

Integer x = new Integer (0);
int i = 1;
System.out.println("i   =  1 -->" + x.toBinaryString( i ));
i <<= 31;
System.out.println("i <<= 31 -->" + x.toBinaryString( i ));
i >>= 31;
System.out.println("i >>= 31 -->" + x.toBinaryString( i ));
i >>= 1;
System.out.println("i >>=  1 -->" + x.toBinaryString( i ));

each output line shows the binary representation of i.
Note, the first line shows 1 as its binary representation because the leading zeros are omitted.
Hope this helped. But basically it has to do with the fact that int has 32-bit size, and the left most bit is the sign bit that also happens to be the padding bit for >>. Remember >>> uses zero as its padding bit regardless of the sign bit.

[bani kaali]

I didnt know about toBinaryString() .Thanks chung for helping me .