Dear all,

In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
integers and y not equal to zero

is there any other identity involving % operator..??? or any other
binary operator..??

for example

w % n = w & (n-1); when n is a power of 2.

sophia wrote:

> Dear all,
>
> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> integers and y not equal to zero

No.

Try x = 5 and y = 3. Straightaway x/y will give you 0.

<snip>

santosh wrote:

> sophia wrote:
>
>> Dear all,
>>
>> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
>> integers and y not equal to zero
>
> No.
>
> Try x = 5 and y = 3. Straightaway x/y will give you 0.
>
> <snip>

Sorry to the OP. This is wrong. Your equation is indeed correct.

On Apr 25, 2:06pm, santosh <santosh....@gmail.com> wrote:
> sophia wrote:
> > Dear all,
>
> > In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> > integers and y not equal to zero
>
> No.
>
> Try x = 5 and y = 3. Straightaway x/y will give you 0.

see this
#include<stdio.h>
#include<stdlib.h>

int main(void)
{
int x,y;
x=5,y=3;

printf("\n%d", ( x -(x/y) * y));
printf("\n%d", x%y);

return EXIT_SUCCESS;
}

sophia wrote:

> On Apr 25, 2:06pm, santosh <santosh....@gmail.com> wrote:
>> sophia wrote:
>> > Dear all,
>>
>> > In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
>> > integers and y not equal to zero
>>
>> No.
>>
>> Try x = 5 and y = 3. Straightaway x/y will give you 0.
>
> see this
> #include<stdio.h>
> #include<stdlib.h>
>
> int main(void)
> {
> int x,y;
> x=5,y=3;
>
> printf("\n%d", ( x -(x/y) * y));
> printf("\n%d", x%y);
>
> return EXIT_SUCCESS;
> }

Yes, my response was incorrect. I'll leave it to the others to answer
your original query since it's meaning is rather unclear to me.

There are one more interesting fact for integers:

x%y==0 <==> (x/y)*y==x


sophia wrote:
> Dear all,
>
> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> integers and y not equal to zero
>
> is there any other identity involving % operator..??? or any other
> binary operator..??
>
> for example
>
> w % n = w & (n-1); when n is a power of 2.
>

santosh wrote:
> sophia wrote:
>
>> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y
>> are integers and y not equal to zero
>
> No. Try x = 5 and y = 3. Straightaway x/y will give you 0.

Huh? I find x/y yields 1.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.

** Posted from http://www.teranews.com **

On Fri, 25 Apr 2008 01:30:05 -0700 (PDT), sophia
<sophia.agnes@gmail.com> wrote:

>Dear all,
>
>In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
>integers and y not equal to zero
>
>is there any other identity involving % operator..??? or any other
>binary operator..??
>
>for example
>
> w % n = w & (n-1); when n is a power of 2.

If your instructor is spending this much time on esoteric
equivalences, you need to find a new one. If you are doing this
yourself, your time would be much better spent on mastering the
fundamental concepts of the language before you go looking for "cute
tricks".


Remove del for email

On Fri, 25 Apr 2008 12:27:18 +0300,
Eugeny Myunster <box@eugeny.ws> wrote:
> sophia wrote:
>> Dear all,
>>
>> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
>> integers and y not equal to zero

> There are one more interesting fact for integers:
>
> x%y==0 <==> (x/y)*y==x

To me, that seems to be the same 'fact'.

Martien
--
|
Martien Verbruggen | We are born naked, wet and hungry. Then
| things get worse.
|

sophia wrote:
> Dear all,
>
> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> integers and y not equal to zero
>
> is there any other identity involving % operator..??? or any other
> binary operator..??
>
> for example
>
> w % n = w & (n-1); when n is a power of 2.

int n_is_Power_of_two(long unsigned n)
{
return (n & n - 1) == 0 && n != 0;
}

int n_is_Power_of_four(long unsigned n)
{
return (n & n - 1) == 0 && n % 3 == 1;
}

int n_is_Power_of_eight(long unsigned n)
{
return (n & n - 1) == 0 && n % 7 == 1;
}

--
pete

"sophia" <sophia.agnes@gmail.com> wrote in message
news:79d97bf1-4071-4637-934e-21428324a1b0@y22g2000prd.googlegroups.com...
> Dear all,
>
> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> integers and y not equal to zero

WTF? x%y does not always equal zero. What use woudl that be?!

sophia <sophia.agnes@gmail.com> writes:

> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> integers and y not equal to zero

Yes, but you have to add also that x/y is representable. When x =
INT_MIN and y = -1, x/y may not be representable.

> is there any other identity involving % operator..??? or any other
> binary operator..??
>
> for example
>
> w % n = w & (n-1); when n is a power of 2.

Again you have to add extra conditions. It requires n > 1 and w >= 0.

--
Ben.

"ArWeGod" <ArWeGod?@sbcglobal.net> writes:

> "sophia" <sophia.agnes@gmail.com> wrote in message
> news:79d97bf1-4071-4637-934e-21428324a1b0@y22g2000prd.googlegroups.com...
>> Dear all,
>>
>> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
>> integers and y not equal to zero
>
> WTF? x%y does not always equal zero. What use woudl that be?!

The expression given is a C arithmetic expression. It is not intended
to be read as mathematics.

--
Ben.

On 27 Jul, 15:29, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> "ArWeGod" <ArWeG...@sbcglobal.net> writes:
> > "sophia" <sophia.ag...@gmail.com> wrote in message
> >news:79d97bf1-4071-4637-934e-21428324a1b0@y22g2000prd.googlegroups.com....

> >> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> >> integers and y not equal to zero
>
> > WTF? x%y does not always equal zero. What use woudl that be?!
>
> The expression given is a C arithmetic expression. It is not intended
> to be read as mathematics.

I'd have said exactly the opposite. It is not a C expression
(because x%y does not make sense on the right hand side of
an assignment), but it *is* a valid mathematical expression
(given a suitable definition of /). I've occaisionally
found it useful in languages that don't provide a remainder
operator.

--
Nick Keighley

A lot of the c.l.c. verbiage seems to be devoted to the numerical
density of cavorting nubile seraphim upon pinheads.
CBFalconer

Nick Keighley wrote:
>
.... snip ...
>
> --
> Nick Keighley
>
> A lot of the c.l.c. verbiage seems to be devoted to the numerical
> density of cavorting nubile seraphim upon pinheads.
> CBFalconer

Did I ever say that? I don't remember it. Seems accurate, though.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.

On 28 Jul, 20:25, CBFalconer <cbfalco...@yahoo.com> wrote:
> Nick Keighleywrote:

> > --
> >Nick Keighley
>
> > A lot of the c.l.c. verbiage seems to be devoted to the numerical
> > density of cavorting nubile seraphim upon pinheads.
> > CBFalconer
>
> Did I ever say that? I don't remember it. Seems accurate, though.

It was from some time ago, but I *believe* I correctly attributed it.

<google>

yep, 2001-03-02
subject: "Sleep command for noninteger time"

--
Nick Keighley

"Resistance is futile. Read the C-faq."
-- James Hu (c.l.c.)

Nick Keighley wrote:
> CBFalconer <cbfalco...@yahoo.com> wrote:
>> Nick Keighleywrote:
>
>>> --
>>> Nick Keighley
>>
>>> A lot of the c.l.c. verbiage seems to be devoted to the numerical
>>> density of cavorting nubile seraphim upon pinheads.
>>> CBFalconer
>>
>> Did I ever say that? I don't remember it. Seems accurate, though.
>
> It was from some time ago, but I *believe* I correctly attributed it.
>
> <google>
>
> yep, 2001-03-02
> subject: "Sleep command for noninteger time"

Oh my. Only 7 years ago, and what a difference. My wife,
brother-in-law, sister-in-law, cat, dog were all alive, I was not
fully retired, hadn't had 3 serious operations, and Bush hadn't had
a chance to do serious harm.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.

"CBFalconer" <cbfalconer@yahoo.com> wrote in message
news:488F732B.2D78A640@yahoo.com...
> Nick Keighley wrote:
> > CBFalconer <cbfalco...@yahoo.com> wrote:
> >> Nick Keighleywrote:
> >
> >>> --
> >>> Nick Keighley
> >>
> >>> A lot of the c.l.c. verbiage seems to be devoted to the numerical
> >>> density of cavorting nubile seraphim upon pinheads.
> >>> CBFalconer
> >>
> >> Did I ever say that? I don't remember it. Seems accurate, though.
> >
> > It was from some time ago, but I *believe* I correctly attributed it.
> >
> > <google>
> >
> > yep, 2001-03-02
> > subject: "Sleep command for noninteger time"
>
> Oh my. Only 7 years ago, and what a difference. My wife,
> brother-in-law, sister-in-law, cat, dog were all alive, I was not
> fully retired, hadn't had 3 serious operations, and Bush hadn't had
> a chance to do serious harm.

Define Bush.

"Ben Bacarisse" <ben.usenet@bsb.me.uk> wrote in message
news:87ej5f764i.fsf@bsb.me.uk...
> "ArWeGod" <ArWeGod?@sbcglobal.net> writes:
>
> > "sophia" <sophia.agnes@gmail.com> wrote in message
> >
news:79d97bf1-4071-4637-934e-21428324a1b0@y22g2000prd.googlegroups.com...
> >> Dear all,
> >>
> >> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
> >> integers and y not equal to zero
> >
> > WTF? x%y does not always equal zero. What use woudl that be?!
>
> The expression given is a C arithmetic expression. It is not intended
> to be read as mathematics.

Okay, I don't see how something arithmetic can avoid being mathematics, but
let's push forward.

All I see is X divided by Y, times Y (i.e. X, for those following along)
subtracted from itself to acheive ZERO in 100% of the cases. So I guess I
don't see how this works...

Let's do the math (or arithmetic):

10 % 10 = 10 - ( (10/10, ie 1) * 10 (ie: back to 10)) = ZERO!!! Yay!
2 % 10 = 2 minus (2 divided by 10 times 10) = ?
Anyone.... Anyone...

--
ArWeConfused

>> >> In C, x % y = ( x -(x/y) * y); is n't it ...???? where x and y are
>> >> integers and y not equal to zero
>> >
>> > WTF? x%y does not always equal zero. What use woudl that be?!
>>
>> The expression given is a C arithmetic expression. It is not intended
>> to be read as mathematics.
>
>Okay, I don't see how something arithmetic can avoid being mathematics, but
>let's push forward.

The / in that expression is INTEGER DIVISION. An integer divided
by an integer yields an integer result.

>All I see is X divided by Y, times Y
>(i.e. X, for those following along)

That's true if you are doing your mathematics in reals, but not if
you're doing it in integers.

>subtracted from itself to acheive ZERO in 100% of the cases. So I guess I
>don't see how this works...

15/10 in C is 1, not 1.5, since the result of dividing two integers
is an integer.

>Let's do the math (or arithmetic):
>
>10 % 10 = 10 - ( (10/10, ie 1) * 10 (ie: back to 10)) = ZERO!!! Yay!
>2 % 10 = 2 minus (2 divided by 10 times 10) = ?
>Anyone.... Anyone...

2/10 is zero.

> 2/10 is zero.

Oh. Right. Carry on.

--
ArWeWrong/ArWeNew=1 (cos it's TRUE!)

On Jul 30, 10:18 am, gordonb.ed...@burditt.org (Gordon Burditt) wrote:
<snip>
> 2/10 is zero.

Please don't remove attributions.

On Jul 30, 12:22 pm, "ArWeGod" <ArWeG...@sbcglobal.net> wrote:
> > 2/10 is zero.
>
> Oh. Right. Carry on.

Please don't remove attributions.

Hello Sophia, list

On Apr 25, 4:30 am, sophia <sophia.ag...@gmail.com> wrote:

http://en.wikipedia.org/wiki/Operato...etic_operators
http://en.wikipedia.org/wiki/Modulo_operation

Regards
Ccello

By the way... CBFalconer, sorry about your wife, brother-in-law,
sister-in-law, cat and dog.

ArWeGod wrote:
>
.... snip ...
>
> Let's do the math (or arithmetic):
>
> 10 % 10 = 10 - ( (10/10, ie 1) * 10 (ie: back to 10)) = ZERO!!! Yay!
> 2 % 10 = 2 minus (2 divided by 10 times 10) = ?
> Anyone.... Anyone...

Using the usual integer arithmetic, 2 / 10 is unequivicably zero.
After which zero times 10 remains zero, and that is the value of
the expression.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.