even or odd

[melsayid]

The program always shows that the input is odd.

int main ()
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);
d=1;
if (d==n)
{
printf ("The number is odd\n");
}
else if (d < n)
{
for (; d<n; d+=2);
printf ("The number is odd\n");
}
else
{
printf ("The number is even\n");
}
--
Mustafa El Sayid

[Richard Bos]

melsayid <me******@gmail.comwrote:

The program always shows that the input is odd.

No, it doesn't. Try entering 0.

You might want to look up the modulo operator, BTW.

Richard

[Morris Dovey]

melsayid wrote:

>
The program always shows that the input is odd.

int main ()
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);
d=1;
if (d==n)
{
printf ("The number is odd\n");
}
else if (d < n)
{
for (; d<n; d+=2);
printf ("The number is odd\n");
}
else
{
printf ("The number is even\n");
}

This should work for twos complement machines:

#include <stdio.h>
int main(void)
{ int n;
printf ("Enter a Number: ");
scanf ("%d", &n);
printf("n is %s\n",(n & 1)?"odd":"even");
return 0;
}

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

[melsayid]

On Mar 19, 4:57 pm, [e-mail] (Richard Bos) wrote:

melsayid <[e-mail]:

The program always shows that the input is odd.

No, it doesn't. Try entering 0.

You might want to look up the modulo operator, BTW.

Richard

Try entering 2 or any higher even number.

--
Mustafa Zaza

[Richard Tobin]

In article <69**********************************@e23g2000prf. googlegroups.com>,
melsayid <me******@gmail.comwrote:

else if (d < n)
{
for (; d<n; d+=2);
printf ("The number is odd\n");
}

What is the point of the loop here? You don't test anything after
it; you just print "The number is odd".

-- Richard
--
:wq

[Richard Tobin]

In article <47***************@iedu.com>,
Morris Dovey <mr*****@iedu.comwrote:

>This should work for twos complement machines:

#include <stdio.h>
int main(void)
{ int n;
printf ("Enter a Number: ");
scanf ("%d", &n);
printf("n is %s\n",(n & 1)?"odd":"even");
return 0;
}

It will work for non-negative ints regardless of whether the machine
uses twos complement.

-- Richard
--
:wq

[melsayid]

On Mar 19, 5:19 pm, [e-mail] (Richard Tobin) wrote:

In article <[e-mail]>,

melsayid <[e-mail]:

else if (d < n)
{
for (; d<n; d+=2);
printf ("The number is odd\n");
}

What is the point of the loop here? You don't test anything after
it; you just print "The number is odd".

-- Richard
--
:wq

I've removed the semicolon and it still doesn't work probably.
Here is how I thought about it,
*The input is n
1-Equals the d with 1
2-Check if the n is equal to d, if yes it prints that the number is
odd
3-If not it checks if the d is less with n, if not it keeps on adding
+2 to the d till it's equal to the value of the n then output that
it's odd
4-Anything else should output that the input is even

@Morris
It works for me, but I want to figure out what's wrong with my code.
Also I don't understand your code, I just started learning C.

--
Mustafa El Sayid

[Richard Tobin]

In article <cc**********************************@e23g2000prf. googlegroups.com>,
melsayid <me******@gmail.comwrote:

for (; d<n; d+=2);
printf ("The number is odd\n");

>3-If not it checks if the d is less with n, if not it keeps on adding
+2 to the d till it's equal to the value of the n then output that
it's odd

Suppose n is 8. You add 2 to d until it's greater than or equal to n.
In this case, that will happen when d is 9. But you still print out
that n is odd! You don't do anything different depending on whether d
stopped at n or went one past it.

-- Richard
--
:wq

[Morris Dovey]

Richard Tobin wrote:

>
In article <47***************@iedu.com>,
Morris Dovey <mr*****@iedu.comwrote:

This should work for twos complement machines:

#include <stdio.h>
int main(void)
{ int n;
printf ("Enter a Number: ");
scanf ("%d", &n);
printf("n is %s\n",(n & 1)?"odd":"even");
return 0;
}

It will work for non-negative ints regardless of whether the machine
uses twos complement.

You noticed that, eh?

:-D

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

[melsayid]

On Mar 19, 5:35 pm, [e-mail] (Richard Tobin) wrote:

In article <[e-mail]>,

melsayid <[e-mail]:

for (; d<n; d+=2);
printf ("The number is odd\n");

3-If not it checks if the d is less with n, if not it keeps on adding
+2 to the d till it's equal to the value of the n then output that
it's odd

Suppose n is 8. You add 2 to d until it's greater than or equal to n.
In this case, that will happen when d is 9. But you still print out
that n is odd! You don't do anything different depending on whether d
stopped at n or went one past it.

-- Richard
--
:wq

Ok, it's working now.
Thanks.

--
Mustafa El Sayid

[Kenneth Brody]

melsayid wrote:

>
The program always shows that the input is odd.

Not "always", just "for any positive number". (See also Richard Bos'
reply.)

int main ()
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);

You should probably check for failure here, but you can get away
without it for simple test cases where you are entering the data.

d=1;
if (d==n)
{
printf ("The number is odd\n");
}
else if (d < n)

This statement will be true for any "n" greater than 1.

{
for (; d<n; d+=2);

Here, you simply increment d by 2 until it is at least equal to n.
Consider, however, how long your program will run if you enter a
number such as 2000000000. Also consider that this may never
terminate if you enter 2147483647.

<pedant>
Note my use of the word "may" in the preceding sentence. :-)
</pedant>

printf ("The number is odd\n");

And then unconditionally print "The number is odd".

}
else

Only zero or negative numbers will ever get here.

{
printf ("The number is even\n");

And then unconditionally print "The number is even".

}

Look up the modulo operator -- % -- as it will suit your needs.

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer.h|
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:Th*************@gmail.com>

[Robbie Hatley]

"melsayid" sed:

The program always shows that the input is odd.

No it doesn't.

It prints
"even" for all numbers < 1
"odd" for all numbers >= 1

Which, of course, makes no sense whatsoever.

int main ()
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);
d=1;
if (d==n)
{
printf ("The number is odd\n");
}
else if (d < n)
{
for (; d<n; d+=2);
printf ("The number is odd\n");
}
else
{
printf ("The number is even\n");
}

Write it in English first. Here's what you just wrote, translated into
plain English:

if d is 1 or greater, print "odd"
else print "even"

Does it make any sense in English? No. So it won't make any sense in C,
either!

Look-up "modulo" in a math book. Then look-up "modulo" in a C book.
(Hint: its the "%" operator.)

Then ask yourself, "what do the words 'even' and 'odd' mean?".
(Hint: it has to do with whether a number is divisible by 2.)

Then ask yourself, "what is the relationship betwen 'divisibility' and
'modulo'?"

I could easily re-write your program for you to correctly print
"odd" or "even", but I won't be a prick and spoil the fun you
could have by forcing yourself to learn these things for yourself.

--
Cheers,
Robbie Hatley
lonewolf aatt well dott com
www dott well dott com slant user slant lonewolf slant

[Kenneth Brody]

melsayid wrote:

>
On Mar 19, 5:35 pm, [e-mail] (Richard Tobin) wrote:

In article <[e-mail]>,

melsayid <[e-mail]:

for (; d<n; d+=2);
printf ("The number is odd\n");
>3-If not it checks if the d is less with n, if not it keeps on adding
>+2 to the d till it's equal to the value of the n then output that
>it's odd

Suppose n is 8. You add 2 to d until it's greater than or equal to n.
In this case, that will happen when d is 9. But you still print out
that n is odd! You don't do anything different depending on whether d
stopped at n or went one past it.

-- Richard
--
:wq

Ok, it's working now.
Thanks.

Could you post your updated "working" code?

Have you considered the ramifications of your logic should the
person enter 2000000000 (or 2147483647) as the number?

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer.h|
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:Th*************@gmail.com>

[Mahesh]

On Mar 20, 6:19*pm, Kenneth Brody <[e-mail]:

melsayid wrote:

On Mar 19, 5:35 pm, [e-mail] (Richard Tobin) wrote:

In article <[e-mail]>,

melsayid *<[e-mail]:
* * * * *for (; d<n; d+=2);
* * * * *printf ("The number is odd\n");
3-If not it checks if the d is less with n, if not it keeps on adding
+2 to the d till it's equal to the value of the n then output that
it's odd

Suppose n is 8. *You add 2 to d until it's greater than or equal to n.
In this case, that will happen when d is 9. *But you still print out
that n is odd! *You don't do anything different depending on whetherd
stopped at n or went one past it.

-- Richard
--
:wq

Ok, it's working now.
Thanks.

Could you post your updated "working" code?

Have you considered the ramifications of your logic should the
person enter 2000000000 (or 2147483647) as the number?

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody * * * *|www.hvcomputer.com| #include * * * * * * *|
| kenbrody/at\spamcop.net |www.fptech.com* * | * *<std_disclaimer.h|
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:[e-mail]>- Hide quoted text -

- Show quoted text -

Lets keep it simple

check for the least significant bit of the number, if its 0 then its
even else its odd.
something like

if(num & 0x1)
printf(" ODD \n");
else
printf(" EVEN \n");

happy coding

[santosh]

Mahesh wrote:

<code to check host endianness>

Lets keep it simple

check for the least significant bit of the number, if its 0 then its
even else its odd.
something like

if(num & 0x1)
printf(" ODD \n");
else
printf(" EVEN \n");

happy coding

But this won't detect mixed-endian machines.

[Eric Sosman]

santosh wrote:

Mahesh wrote:

<code to check host endianness>

>Lets keep it simple

check for the least significant bit of the number, if its 0 then its
even else its odd.
something like

if(num & 0x1)
printf(" ODD \n");
else
printf(" EVEN \n");

happy coding

But this won't detect mixed-endian machines.

Endianness doesn't affect the outcome.

Where it fails is with ones'-complement negative values:
it would diagnose -1 as even and -2 as odd. It might be
possible to generate two numbers X and Y such that X is
reported as even and Y as odd, but X==Y is true.

--
Er*********@sun.com

[pete]

melsayid wrote:

>
The program always shows that the input is odd.

scanf ("%d", &n);
d=1;

That's because (d=1) is always odd.

--
pete

[pete]

pete wrote:

>
melsayid wrote:

The program always shows that the input is odd.

scanf ("%d", &n);
d=1;

That's because (d=1) is always odd.

But that's not it. I read it wrong.

It's because this part of the code always prints
"The number is odd\n"
regardless of what happens in the loop.

else if (d < n)
{
for (; d<n; d+=2);
printf ("The number is odd\n");
}

This is my version of the repaired algorithm:

/* BEGIN new.c */

#include <stdio.h>

int main(void)
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);
d = 1;

if (n d) {
do {
d += 2;
} while (n d);
} else {
while (d n) {
n += 2;
}
}
if (d == n) {
puts("The number is odd.");
} else {
puts("The number is even.");
}
return 0;
}

/* END new.c */
--
pete

[Kenneth Brody]

pete wrote:
[...]

This is my version of the repaired algorithm:

/* BEGIN new.c */

#include <stdio.h>

int main(void)
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);
d = 1;

if (n d) {
do {
d += 2;
} while (n d);
} else {
while (d n) {
n += 2;
}
}
if (d == n) {
puts("The number is odd.");
} else {
puts("The number is even.");
}
return 0;
}

/* END new.c */

How long will the program take to run if you enter "2000000000"?
What about "2147483647" on a 32-bit int system?

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer.h|
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:Th*************@gmail.com>

[pete]

Kenneth Brody wrote:

>
pete wrote:
[...]

This is my version of the repaired algorithm:

/* BEGIN new.c */

#include <stdio.h>

int main(void)
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);
d = 1;

if (n d) {
do {
d += 2;
} while (n d);
} else {
while (d n) {
n += 2;
}
}
if (d == n) {
puts("The number is odd.");
} else {
puts("The number is even.");
}
return 0;
}

/* END new.c */

How long will the program take to run if you enter "2000000000"?
What about "2147483647" on a 32-bit int system?

You can't speed up a program that doesn't tell you
anything that you don't already know,
to point where the program is fast enough to become useful.

--
pete

[Richard Heathfield]

John H. Guillory said:

<snip>

Not to
mention, most C compilers have a EVEN() and ODD() macro

They do?

--
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

[Gilbert Hamm]

"John H. Guillory" <jo***@communicomm.comwrote in message
news:tn********************************@4ax.com...

On Wed, 19 Mar 2008 07:52:22 -0700 (PDT), melsayid
<me******@gmail.comwrote:

>>The program always shows that the input is odd.

int main ()
{
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);

if ((n & 1) == 1) printf("The number is odd");
}

Anytime Bit 1 is set, the number is odd, if its cleared, its even.
Simply test for bit 1 and you know if its even or odd.... Not to
mention, most C compilers have a EVEN() and ODD() macro

Why don't you try
if(N%2 == 0) printf("even");
else printf("odd");

[CBFalconer]

"John H. Guillory" wrote:

melsayid <me******@gmail.comwrote:

>The program always shows that the input is odd.

int main () {
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);

if ((n & 1) == 1) printf("The number is odd");

>}

Anytime Bit 1 is set, the number is odd, if its cleared, its
even. Simply test for bit 1 and you know if its even or odd.
Not to mention, most C compilers have a EVEN() and ODD() macro

You should make n an unsigned int. It won't work with signed ints
on 1's complement systems.

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

[Richard Heathfield]

CBFalconer said:

"John H. Guillory" wrote:

>melsayid <me******@gmail.comwrote:

>>The program always shows that the input is odd.

int main () {
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);

if ((n & 1) == 1) printf("The number is odd");

>>}

Anytime Bit 1 is set, the number is odd, if its cleared, its
even. Simply test for bit 1 and you know if its even or odd.
Not to mention, most C compilers have a EVEN() and ODD() macro

You should make n an unsigned int. It won't work with signed ints
on 1's complement systems.

Even on ones' complement systems, it will work with 50% of signed ints.

--
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

[Scott Fluhrer]

"Richard Heathfield" <rj*@see.sig.invalidwrote in message
news:u_******************************@bt.com...

CBFalconer said:

>"John H. Guillory" wrote:

>>melsayid <me******@gmail.comwrote:

The program always shows that the input is odd.

int main () {
int n, d;

printf ("Enter a Number: ");
scanf ("%d", &n);
if ((n & 1) == 1) printf("The number is odd");
}

Anytime Bit 1 is set, the number is odd, if its cleared, its
even. Simply test for bit 1 and you know if its even or odd.
Not to mention, most C compilers have a EVEN() and ODD() macro

You should make n an unsigned int. It won't work with signed ints
on 1's complement systems.

Even on ones' complement systems, it will work with 50% of signed ints.

Well,

if (1) printf("The number is odd");

also works with about 50% of signed ints :-)

--
poncho

[Richard Heathfield]

Scott Fluhrer said:

>
"Richard Heathfield" <rj*@see.sig.invalidwrote in message
news:u_******************************@bt.com...

>CBFalconer said:

>>"John H. Guillory" wrote:
melsayid <me******@gmail.comwrote:

The program always shows that the input is odd.
>
int main () {
int n, d;
>
printf ("Enter a Number: ");
scanf ("%d", &n);
if ((n & 1) == 1) printf("The number is odd");
}

Anytime Bit 1 is set, the number is odd, if its cleared, its
even. Simply test for bit 1 and you know if its even or odd.
Not to mention, most C compilers have a EVEN() and ODD() macro

You should make n an unsigned int. It won't work with signed ints
on 1's complement systems.

Even on ones' complement systems, it will work with 50% of signed ints.

Well,

if (1) printf("The number is odd");

also works with about 50% of signed ints :-)

I think we have a winner!

Of course, my point was that it would have been better to say not that "it
won't work with signed ints on 1's complement systems" but that "it won't
work with *negative* ints on ones' complement systems".

And that's assuming you can /find/ a ones' complement system.

--
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