shiftoperators

[chandana]

anyone please explain the following program
The following code will print
1: int i = 1;
2: i <<= 31;<br /> 3: i >>= 31;
4: i >>= 1;
5:
6: int j = 1;
7: j <<= 31;<br /> 8: j >>= 31;
9:
10: System.out.println("i = " +i );
11: System.out.println("j = " +j);
A) i = 1
j = 1
B) i = -1
j = 1

C) i = 1
j = -1

D) i = -1
j = -1

answer
thanks in advance

[bill bozeman]

chandana,
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Thanks
Bill

[bill bozeman]

Now to answer your question, the answer is D.
Take 1 and do a left shift 31 spaces and you get 100000000...
Now, do a signed right shift by 31 and you get 111111111....
Now do a signed right shift by 1 more and you get 11111111... still
Which leaves you with -1. Same for j but you don't do the last step, but since the last step doesn't do anything, it doesn't really matter.
Bill

[Bala Arul]

Answers : -1, -1.
After 1 is left shifted by 31, the high order bit is 1. When this is right shifted by 31, all the 32 bits are 1's. That is -1. In case of i which is again right shifted by 1, still all the bits are 1's. Thus the result.
Bala Arul.