1>>x behaviour

[Vivien Parlat]

Hello,

I have a little question about operator >between integers: does
anyone know why, when I write:
"1 >x", the returned value is 1 for all values multiple of 32 and 0
else (this does not hurt me, 1 does).
When I do the same thing with : "1L >x", only multiples of 64 do
this.
I thought 0 was the only possible result for any x 0.
Why is it not the case ? And how to correct this simply -- in one
instruction, else I did it by:

long longFoo = intBar;
longFoo >>= x;
return (int) longFoo;

.... but this looks weird to do this.

Thanks in advance to any answer.

[Morten Wennevik]

Hi Vivien,

The shift operation is done using the lowest 5 bits of the x operand for
Int32, and lowest 6 bits for Int64.

1 >32 is the same as 1 >0
On Thu, 23 Nov 2006 12:20:55 +0100, Vivien Parlat <po****@gmail.c omwrote:

Hello,

I have a little question about operator >between integers: does
anyone know why, when I write:
"1 >x", the returned value is 1 for all values multiple of 32 and 0
else (this does not hurt me, 1 does).
When I do the same thing with : "1L >x", only multiples of 64 do
this.
I thought 0 was the only possible result for any x 0.
Why is it not the case ? And how to correct this simply -- in one
instruction, else I did it by:

long longFoo = intBar;
longFoo >>= x;
return (int) longFoo;

... but this looks weird to do this.

Thanks in advance to any answer.

--
Happy Coding!
Morten Wennevik [C# MVP]

[Jay]

So is that a bug?

"Morten Wennevik" <Mo************ @hotmail.comwro te in message
news:op******** *******@tr024.b ouvet.no...
Hi Vivien,

The shift operation is done using the lowest 5 bits of the x operand for
Int32, and lowest 6 bits for Int64.

1 >32 is the same as 1 >0
On Thu, 23 Nov 2006 12:20:55 +0100, Vivien Parlat <po****@gmail.c omwrote:

Hello,

I have a little question about operator >between integers: does
anyone know why, when I write:
"1 >x", the returned value is 1 for all values multiple of 32 and 0
else (this does not hurt me, 1 does).
When I do the same thing with : "1L >x", only multiples of 64 do
this.
I thought 0 was the only possible result for any x 0.
Why is it not the case ? And how to correct this simply -- in one
instruction, else I did it by:

long longFoo = intBar;
longFoo >>= x;
return (int) longFoo;

... but this looks weird to do this.

Thanks in advance to any answer.

--
Happy Coding!
Morten Wennevik [C# MVP]

[Morten Wennevik]

On Fri, 24 Nov 2006 10:14:50 +0100, Jay <nospamwrote:

So is that a bug?

It's by design. When you think about it, shifting a 32 bit integer with
anything more than 31 places doesn't make any sense.

http://msdn2.microsoft.com/en-us/library/xt18et0d.aspx
--
Happy Coding!
Morten Wennevik [C# MVP]

[garyusenet]

What does shift do? I read this thread and was intrigued but don't
understand it's use.
I am just learning the basics.

Please advise.

Morten Wennevik wrote:

On Fri, 24 Nov 2006 10:14:50 +0100, Jay <nospamwrote:

So is that a bug?

It's by design. When you think about it, shifting a 32 bit integer with
anything more than 31 places doesn't make any sense.

http://msdn2.microsoft.com/en-us/library/xt18et0d.aspx
--
Happy Coding!
Morten Wennevik [C# MVP]

[Marc Gravell]

But there is an oddity here;

shifting by 5 and then by 7 is equivalent to shifting by 12
shifting by 16 and then by 16 would leave nothing; yet shifting by 32 in one
go doesn't always...

Intriguing, but not a huge problem generally...

Marc

[Marc Gravell]

In brief, it works on the bitwise representation of a value, and moves
everything left or right by {x} places, dropping bits off one end, and
back-filling with 0s at the other.

e.g. integer 1 in binary is ...00001 (=int 1); left shift gives ...00010
(=int 2); right shift this twice yields ...00000 (=int 0)

http://msdn.microsoft.com/library/de...ftoperator.asp

Marc

[Morten Wennevik]

Hi,

Shifting is used when you need information at bitlevel

Consider the number 68437143. In the file it would be represented as

00000100 00010100 01000100 10010111

Hidden inside is the dimensions of an image where the information is
arranged as

0000 010000 01010001000 10010010111
Unused Colors Width Height
4 6 11 11

To be able to get this information we need to do bit manipulation
By filtering the number with 2047 we get the lower 11 bits

000001000001010 001000100100101 11 (68437143)
&
000000000000000 000000111111111 11 (2047)
=
000000000000000 000000100100101 11 (1175)
[code]
int n = 68437143;
int filter = 2047
int Height = n & filter; // 1175
To get Width we shift the bits 11 places to the right

000001000001010 001000100100101 11 (68437143)

>11

000000000000000 010000010100010 00 (33416) 10010010111 has been pushed out

[code]
n = n >11;

Do the same filtering to get the Width

000000000000000 010000010100010 00 (33416)
&
000000000000000 000000111111111 11 (2047)
=
000000000000000 000000010100010 00 (648)

Shift right 11 more places

000000000000000 010000010100010 00 (33416)

>11

000000000000000 000000000000100 00 (16)

Summarized in code it will look like

int number = 68437143;
int filter = 2047;
int height = number & filter;
number = number >11; // you could also write number >>= 11
int width = number & filter;
int color = number >11;

Image size = 648x1175x16

For instance, image files pack values int bits, not bytes.

On Fri, 24 Nov 2006 10:44:45 +0100, <ga********@myw ay.comwrote:

What does shift do? I read this thread and was intrigued but don't
understand it's use.
I am just learning the basics.

Please advise.

Morten Wennevik wrote:

>On Fri, 24 Nov 2006 10:14:50 +0100, Jay <nospamwrote:

So is that a bug?

It's by design. When you think about it, shifting a 32 bit integer with
anything more than 31 places doesn't make any sense.

http://msdn2.microsoft .com/en-us/library/xt18et0d.aspx
--
Happy Coding!
Morten Wennevik [C# MVP]

--
Happy Coding!
Morten Wennevik [C# MVP]

[Jay]

Yes it doesn't make sense, but it ought to do something sensible rather than something quirky. For
example, if you shift by more than 32 bits, it should set the variable to zero (assuming unsigned
variables for now) since this is the true answer if you did shift more than 32 bits.

"Morten Wennevik" <Mo************ @hotmail.comwro te in message
news:op******** *******@tr024.b ouvet.no...
On Fri, 24 Nov 2006 10:14:50 +0100, Jay <nospamwrote:

So is that a bug?

It's by design. When you think about it, shifting a 32 bit integer with
anything more than 31 places doesn't make any sense.

http://msdn2.microsoft.com/en-us/library/xt18et0d.aspx
--
Happy Coding!
Morten Wennevik [C# MVP]

[Morten Wennevik]

On Fri, 24 Nov 2006 14:22:45 +0100, Jay <nospamwrote:

Yes it doesn't make sense, but it ought to do something sensible rather
than something quirky. For
example, if you shift by more than 32 bits, it should set the variable
to zero (assuming unsigned
variables for now) since this is the true answer if you did shift more
than 32 bits.

Not if there is more information in the shift parameter than should be
used for shifting. For instance, if you have a value that needs to be
shiftet several times each time a different number of places. You could
then store all the shift values in a single integer and shift the shift
value n places each shift.

int someNumber = 12345667;
int shiftNumber = 12345;

int value1 = someNumber;
someNumber >>= shiftNumber;
shiftNumber >>= 5;
int value2 = someNumber;
someNumber >>= shiftNumber
shiftNumber >>= 5
....

I can't see much reason to use it, but there may be some performance to be
gained in a heavy duty loop.
--
Happy Coding!
Morten Wennevik [C# MVP]