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Shift operator


Joined: May 05, 2004
Posts: 2
Hey All,
Here is a question found in a java test.
The following code will print
1: int i = 1;
2: i <<= 31;
3: i >>= 31;
4: i >>= 1;
6: int j = 1;
7: j >>= 31;
8: j >>= 31;
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 = 0
I thought the answer was i=0, j=1. But there was no choice like that.
The answer is D.
Can anybody give me an explanation?
C. Nimo
Ranch Hand

Joined: Mar 23, 2004
Posts: 82
it's all about size and representation.
an int value is represented with 32bits. however, it is represented by a notation called 2's complement. basically, this means that if you have a '1' in the leftmost bit of a number, this is a negative number.
now, i=1 means that it is represented by 31 consecutive zeros, and then a '1'. if you shift that 31 bits to the left, you end up with a '1' at the leftmost bit, which makes it a negative number.
the >> shift right, copies the leftmost bit (the 'sign' bit) - so, when you later shift your number 31 bits to the right, you get a binary represantation of all '1'. this is actually -1 in 2's complement notation.
the last shift doesn't matter, since all the bits are 1 anyway, and the >> shift copies the sign bit. so you still have 32 bits, all of them '1'.
now about the j var - this is initialized to 31 bits all zeros, and a 1 at the rightmost bit. the >> of 31 bits drops the 1 over the edge, and you get the binary representation of all zeros. any shift on that will leave you again with the value of zero.

Joined: May 05, 2004
Posts: 2
Thanks Nimo,
i got the "j-part". But i am still having doubts about the "i-part". What do you mean by >> operator copies the leftmost bit. I thought >> operator moves all the bits by one position to the right.
So if i= 1 which means i = 0000 0000 0000 0000 0000 0000 0000 0001 in binary, after doing i <<= 31,
i=1000 0000 0000 0000 0000 0000 0000 0000
Now if I do i >>= 1;
I thought the value of i should be 0100 0000 0000 0000 0000 0000 0000 0000
Can you clarify it?
Dirk Schreckmann

Joined: Dec 10, 2001
Posts: 7023
Twinkle Twinkle,
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C. Nimo
Ranch Hand

Joined: Mar 23, 2004
Posts: 82
There are two kinds of shift-right operators in Java - signed, and unsigned.
a signed shift, which is denoted by >>, shifts the bits right, keeping the sign. the unsigned shift-right, denoted by >>>, inserts zeros on the left.
now, when you have a negative number, which means that it has a leftmost bit with the value of '1' - a signed shift-right will not insert zeros on the left, it will insert '1's.
I agree. Here's the link:
subject: Shift operator