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Help with -ve shift

Vicky Nag
Ranch Hand
Posts: 40
Can anybody please help me in understanding the output below. I am trying some sample programs with shift operators.
class Test
{
public static void main(String args[])
{
System.out.println("Left by -1 = " + (8 << -1));
System.out.println("Right by -1= " + (8 >> -1));
System.out.println("Unsigned Right by -1 = " + (8 >>> -1));
System.out.println("");
System.out.println("Left by -2 = " + (8 << -2));
System.out.println("Right by -2= " + (8 >> -2));
System.out.println("Unsigned Right by -2 = " + (8 >>> -2));
System.out.println("");
System.out.println("Left by -5 = " + (8 << -5));
System.out.println("Right by -5= " + (8 >> -5));
System.out.println("Unsigned Right by -5 = " + (8 >>> -5));
}
}
Output:
Left by -1 = 0
Right by -1= 0
Unsigned Right by -1 = 0
Left by -2 = 0
Right by -2= 0
Unsigned Right by -2 = 0
Left by -5 = 1073741824
Right by -5= 0
Unsigned Right by -5 = 0

Vicky

Ray Stojonic
Ranch Hand
Posts: 326
Hi Vicky,
I found this in the JLS at�15.19:
If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (�15.22.1) with the mask value 0x1f. The shift distance actually used is therefore always in the range 0 to 31, inclusive.

According to the last part of the paragragh, by using -1 etc, you're out of range.
By the rest of the paragragh, here's my best guess at what's happening:
8 << -1;
8 will be an int, so only the rightmost 5 bits of -1 will be used
1 = 00001
1s complement = 11110
2s complement = 11111
so we have 11111 = -1
apply the & operator with a mask of 1f (or 11111)
11111
&11111
______
11111
as an unsigned number 11111 = 31.
so by "8 << -1" you're asking for a left shift of 31 places.
int 8 = 00000000 00000000 00000000 00001000 (32 bits)
Bits shifted off either end are lost, so you're left with a bunch of 0s.
You could >> a negative number, by -1, then you'd fill the entire thing with the sign bit (which is 1 in a negative number) resulting in -1.
"8 << -1" looks like your saying "shift 8 left 1 backwards", if that's really what you want, use 8 >> 1.
Last note: You got output on 8 << -5 because -5 will resolve to 11, so the one 'on bit' didn't shifted off the int.
hth
[ August 15, 2003: Message edited by: Ray Stojonic ]

Vicky Nag
Ranch Hand
Posts: 40
Ray,
Thanx for the detailed explanation.
I now got the concept.
Vicky