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left shift operation

 
vijay gardiner
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x=1;
x << 29 yields 536870912, whose binary representation is all zeroes except '1' in the 30th higher order bit.
x << 30 yields 1073741824, whose binary representation is all zeroes except '1' in the 31th higher order bit.
What happens when x << 31 ?? Can anyone explain please in detail with the regd to the bits in binary form ??
Thanx,
Vijay
 
Jennifer Wallace
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Integers in Java are signed. So the Most Significant Bit is the sign bit.
Here, I am just showing the Binary representation for the integers considered. U cud also get it with the static function of Integer class.... Integer.toBinaryString(int i);
For example,
0000 0000 0000 0000 0000 0000 0000 0010 = 2

1111 1111 1111 1111 1111 1111 1111 1110 = -2
-2 is here represented in two's complement form.
To get to the exact representation of any number in 2's complement notation u cud do it this way,
0000 0000 0000 0000 0000 0000 0000 0010 = 2
Invert it,
1111 1111 1111 1111 1111 1111 1111 1101
Then add 1 to it, +1
1111 1111 1111 1111 1111 1111 1111 1110 = -2

Here,in ur example,initially,i is 1
0000 0000 0000 0000 0000 0000 0000 0001 = 1
then i<<29 is
0010 0000 0000 0000 0000 0000 0000 0000
which is Big Positive number.
i<<31 moves i to the Most significant bit
1000 0000 0000 0000 0000 0000 0000 0000
which is the representation for -2147483648

Hope this helps!
[ February 18, 2002: Message edited by: Jennifer Wallace ]
 
vijay gardiner
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thanx!
that helps!
Vijay
 
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