x % y = ( x -(x/y) * y)

[sophia]

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]

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]

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.

[sophia]

On Apr 25, 2:06*pm, santosh <[e-mail]:

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;
}

[santosh]

sophia wrote:

On Apr 25, 2:06*pm, santosh <[e-mail]:

>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.

[Eugeny Myunster]

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.

[Barry Schwarz]

On Fri, 25 Apr 2008 01:30:05 -0700 (PDT), sophia
<so**********@gmail.comwrote:

>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

[pete]

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

[CBFalconer]

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 **

[ArWeGod]

"sophia" <so**********@gmail.comwrote in message
news:79**********************************@y22g2000 prd.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?!

[Ben Bacarisse]

sophia <so**********@gmail.comwrites:

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.

[Ben Bacarisse]

"ArWeGod" <Ar******@sbcglobal.netwrites:

"sophia" <so**********@gmail.comwrote in message
news:79**********************************@y22g2000 prd.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.

[Nick Keighley]

On 27 Jul, 15:29, Ben Bacarisse <[e-mail]:

"ArWeGod" <[e-mail]:

"sophia" <[e-mail] in message
news:79**********************************@y22g2000 prd.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

[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.

[Nick Keighley]

On 28 Jul, 20:25, CBFalconer <[e-mail]:

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.)

[CBFalconer]

Nick Keighley wrote:

CBFalconer <[e-mail]:

>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.

[ArWeGod]

"CBFalconer" <cb********@yahoo.comwrote in message
news:48***************@yahoo.com...

Nick Keighley wrote:

CBFalconer <[e-mail]:

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.
;)

[ArWeGod]

"Ben Bacarisse" <be********@bsb.me.ukwrote in message
news:87************@bsb.me.uk...

"ArWeGod" <Ar******@sbcglobal.netwrites:

"sophia" <so**********@gmail.comwrote in message

news:79**********************************@y22g2000 prd.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

[Gordon Burditt]

>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.

[ArWeGod]

2/10 is zero.

Oh. Right. Carry on.

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

[vippstar]

On Jul 30, 10:18 am, [e-mail] (Gordon Burditt) wrote:
<snip>

2/10 is zero.

Please don't remove attributions.

[vippstar]

On Jul 30, 12:22 pm, "ArWeGod" <[e-mail]:

2/10 is zero.

Oh. Right. Carry on.

Please don't remove attributions.

[soscpd]

Hello Sophia, list

On Apr 25, 4:30 am, sophia <[e-mail]:

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.

[Barry Schwarz]

On Wed, 30 Jul 2008 00:07:31 -0700, "ArWeGod" <Ar******@sbcglobal.net>
wrote:

>"Ben Bacarisse" <be********@bsb.me.ukwrote in message
news:87************@bsb.me.uk...

>"ArWeGod" <Ar******@sbcglobal.netwrites:

"sophia" <so**********@gmail.comwrote in message

news:79**********************************@y22g200 0prd.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.

Then you are really going to hate
x = x + 1;

--
Remove del for email

[Richard Heathfield]

Barry Schwarz said:

On Wed, 30 Jul 2008 00:07:31 -0700, "ArWeGod" <Ar******@sbcglobal.net>
wrote:

>>"Ben Bacarisse" <be********@bsb.me.ukwrote in message
news:87************@bsb.me.uk...

<snip>

>>>
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.

Then you are really going to hate
x = x + 1;

I don't see why.

x = x + 1 (1)
= 1/x = 1/(x + 1) (2)
= (x + 1) / x = 1 (3)

Subtracting x from both sides of (1), x - x = 1. Substituting into (3):

(x + 1) / x = x - x (4)
= x + 1 = x(x - x) (5)
= x + 1 = x^2 - x^2 (6)

Let y = x^2 (7)
= x + 1 = y - y (8)
= x + 1 = 0 (9)

Subtracting 1 from both sides:

x = -1 (10)

Easy!

;-)

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[CBFalconer]

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.

[CBFalconer]

soscpd wrote:

>

.... snip ...

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

Thanks. I'm gradually realizing I'm probably not immortal.

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

[ArWeGod]

Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

-ArWeNewHere

[ArWeGod]

"ArWeGod" <Ar******@sbcglobal.netwrote in message
news:%M*******************@nlpi064.nbdc.sbc.com...

Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

-ArWeNewHere

Actually, I don't think I'll be back. This isn't what I thought it was. I
think what I want is a .NET group, not this one. This seems to be about
arguing about stuff from 20 years ago. I just want to learn to open some
files and print stuff. The printing of stuff seems even harder than posting
here, but if I get it to work, I will have spent my time better-er.

Peace.

--
ArWeInOz-TNPLH,TNPLH!

[santosh]

ArWeGod wrote:

>Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

Do you see the "ArWeGod wrote:" line above? That's an "attribution"
line. Similar lines should be automatically generated for each poster
whose text appears in a post, by just about every newsreader I'm aware
of, including Google Groups.

If on the off chance that yours doesn't then you should, if possible,
add them manually. They are not *absolutely* essential, but most
respondents will probably killfile you if you continue to post
unattributed quotes.

-ArWeNewHere

And the sig separator is usually of the form "-- \n" (as a C string).
Anything else is prone to be unrecognised as yours was.

[santosh]

ArWeGod wrote:

"ArWeGod" <Ar******@sbcglobal.netwrote in message
news:%M*******************@nlpi064.nbdc.sbc.com...

Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

-ArWeNewHere

Actually, I don't think I'll be back. This isn't what I thought it
was.

And what was that?

I think what I want is a .NET group, not this one. This seems to
be about arguing about stuff from 20 years ago.

C is still widely used. It's use these days is largely "invisible" to
desktop application programmers, that's true, but I suspect that the
volume of new C code continues to be in the top five for all
programming languages.

<snip>

[Nick Keighley]

On 31 Jul, 07:30, "ArWeGod" <[e-mail]:

Actually, I don't think I'll be back. This isn't what I thought it was. I
think what I want is a .NET group, not this one. This seems to be about
arguing about stuff from 20 years ago. I just want to learn to open some
files and print stuff.

C does this

The printing of stuff seems even harder than posting
here, but if I get it to work, I will have spent my time better-er.

printing is pretty easy in C.
#include <stdlib.h>
#include <stdio.h>

int main (void)
{
FILE *in;
char line[80];

/* open a file */
in = fopen ("fred.dat", "r");

if (in == NULL)
return EXIT_FAILURE; /* file didn't open */

/* read a line */
fgets (line, sizeof line, in);

/* close file */
fclose (in);

/* print stuff */
printf ("stuff\n");

return EXIT_SUCCESS;
}

--
Nick Keighley

in comp.lang.c, the very people most capable of making the inference
are those least likely to make it. This newsgroup considers pedantry
to be an art form.
Richard Heathfield

[vippstar]

On Jul 31, 9:15 am, "ArWeGod" <[e-mail]:

Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

It means you're a troll and ignored from now on.

[CBFalconer]

ArWeGod wrote:

>

>Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

Do you see that initial line "ArWeGod wrote;"? My newsreader put
that there when I hit the reply command. It identifies who wrote
what in the quoted section. It has one less '>' leading it than
the material written by the attributee. Don't remove it if you are
quoting anything written by that attributee.

Try the following informative links:
<http://www.catb.org/~esr/faqs/smart-questions.html>
<http://www.caliburn.nl/topposting.html>
<http://www.netmeister.org/news/learn2quote.html>

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

[CBFalconer]

vi******@gmail.com wrote:

"ArWeGod" <[e-mail]:

>>Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

It means you're a troll and ignored from now on.

And this answer means you are an idiot and useless to newbies.

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

[Richard]

CBFalconer <cb********@yahoo.comwrites:

vi******@gmail.com wrote:

>"ArWeGod" <[e-mail]:

>>>Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

It means you're a troll and ignored from now on.

And this answer means you are an idiot and useless to newbies.

*splutter*!

[Kenny McCormack]

In article <g6**********@registered.motzarella.org>,
Richard <rg****@gmail.comwrote:

>CBFalconer <cb********@yahoo.comwrites:

>vi******@gmail.com wrote:

>>"ArWeGod" <[e-mail]:

Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

It means you're a troll and ignored from now on.

And this answer means you are an idiot and useless to newbies.

*splutter*!

Everybody loves a good chick fight!

[Richard]

ga*****@xmission.xmission.com (Kenny McCormack) writes:

In article <g6**********@registered.motzarella.org>,
Richard <rg****@gmail.comwrote:

>>CBFalconer <cb********@yahoo.comwrites:

>>vi******@gmail.com wrote:
"ArWeGod" <[e-mail]:

>Please don't remove attributions.
>
Yeah. Sure. Ummm... what does that mean?

It means you're a troll and ignored from now on.

And this answer means you are an idiot and useless to newbies.

*splutter*!

Everybody loves a good chick fight!

Vippstar seems to have taken Chuck's role as Heathfields favourite
lackey.

We've seen "Indeed"s a plenty. A few "Chapter and Verse" and now the
ritual culling of anyone who doesn't worship the regs immediately.

Could be some good cat fights coming up to lighten the summer.

[vippstar]

On Jul 31, 4:50 pm, CBFalconer <[e-mail]:

[e-mail] wrote:

"ArWeGod" <[e-mail]:

>Please don't remove attributions.

Yeah. Sure. Ummm... what does that mean?

It means you're a troll and ignored from now on.

And this answer means you are an idiot and useless to newbies.

No, I am correct that he is a troll. What *your* reply means, is that
you obviously miss a post or two.
"ArWeGod" only snipped the attributes when he replied to my message;
he did not snip attributes in latter or former messages.

[David Thompson]

On Sun, 27 Jul 2008 15:28:05 +0100, Ben Bacarisse
<be********@bsb.me.ukwrote:

sophia <so**********@gmail.comwrites:

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.

n >= 1. Which all powers of an integer are usually taken to be.

- formerly david.thompson1 || achar(64) || worldnet.att.net

[David Thompson]

On Mon, 28 Jul 2008 01:38:54 -0700 (PDT), Nick Keighley
<ni******************@hotmail.comwrote:

On 27 Jul, 15:29, Ben Bacarisse <[e-mail]:

"ArWeGod" <[e-mail]:

"sophia" <[e-mail] in message
>news:79**********************************@y22g200 0prd.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

<coughleft </>

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.

- formerly david.thompson1 || achar(64) || worldnet.att.net