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bits required


count the bits required to be altered while swaping values a and b
rajm2@gmail.com said:

> count the bits required to be altered while swaping values a and b

Why?

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

rajm2@gmail.com wrote:
> count the bits required to be altered while swaping values a and b

You can repost it all you want, we still won't do your homework for
you.

Back in the day, when I was a teaching assistant (Physics), I used to
tell the students, "It's pointless to cheat on homework. Homework is
for you. We only collect and grade it so you'll get feedback on what
you're doing wrong and to give you incentive to actually do the work.
Exams are where we find out what you really know."

Brian

In article <1180195883.579214.304@a26g2000pre.googlegroups.com>,

 <rajm2@gmail.com> wrote:
>count the bits required to be altered while swaping values a and b

The answer may be ill-defined. If a temporary value is used,
the the number of bits required to be altered to set that
temporary to a certain value would depend on what was in
the temporary to start with, which is unknown if the temporary
is in automatic storage.

There is also the question of how to count the alteration of bits.
If a particular bit gets altered twice, then is that a count of 2
(two alterations), or a count of 1 (one bit altered) ?

Is the requirement to come up with a formula that expresses the
number of bits required to be altered to swap two values? Or is
the requirement to come up with a program that would count the
bits required to be altered to swap two values that are inputs?
Or are the two values given ahead of time?

What do we know about the types of the two values a and b?
Are they integral types? Are they potentially floating point
numbers? Are they potentially of mixed type, one integral
and one floating point?

What are we to do about the problem that "bits" are a matter
of representation (which is not completely defined in C),
whereas "values" are independant of representation? For example,
the answer might be different if the C implementation happens
to be using two's complement to represent values when the programmer
is expecting one's complement representation instead.
--
Programming is what happens while you're busy making other plans.

On 26 May 2007 09:11:23 -0700, rajm2@gmail.com wrote:

>count the bits required to be altered while swaping values a and b

Posting the same question every hour will not get you more or faster
responses.  However, it will increase the number of killfiles that
contain you name.

Posting an obvious homework question while making no attempt to solve
it yourself merely hastens the process.

Remove del for email

Barry Schwarz said:

> On 26 May 2007 09:11:23 -0700, rajm2@gmail.com wrote:

>>count the bits required to be altered while swaping values a and b

> Posting the same question every hour will not get you more or faster
> responses.  However, it will increase the number of killfiles that
> contain you name.

Indeed. In fact, from over here it's already too late.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

rajm2@gmail.com wrote:
> count the bits required to be altered while swaping values a and b

42.7
--
Flash Gordon

rajm2@gmail.com writes:
> count the bits required to be altered while swaping values a and b

Give us your instructor's e-mail address so we can submit the solution
directly.

Optionally, you can also give us your name so we can tell your
instructor who should get the "credit".

Or you might just consider doing your own homework.

--
Keith Thompson (The_Other_Keith) k@mib.org  <http://www.ghoti.net/~kst>
San Diego Supercomputer Center             <*>  <http://users.sdsc.edu/~kst>
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"

rajm2@gmail.com wrote:

> count the bits required to be altered while swaping values a and b

/* BEGIN new.c */

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

unsigned bit_count(unsigned n);

int main(void)
{
    unsigned a, b, mask;

    a = rand();
    b = rand();
    mask = a ^ b;
    printf("a is %u\nb is %u.\n", a, b);
    printf("%u bits would need to be flipped in each object\n"
        "to swap their values.\n", bit_count(mask));
    printf("%u ^ %u is %u\n", a, mask, a ^ mask);
    printf("%u ^ %u is %u\n", b, mask, b ^ mask);
    return 0;

}

unsigned bit_count(unsigned n)
{
    unsigned count;

    for (count = 0; n != 0; n &= n - 1) {
        ++count;
    }
    return count;

}

/* END new.c */

--
pete

On May 27, 3:52 am, pete <pfil@mindspring.com> wrote:

Why you are using unsigned int everywhere?Why count is unsigned?And
why you XOR the inputs?

Bitwise operations are better suited to unsigned types.

> And why you XOR the inputs?

The result of a ^ b,
is a mask which has the bits set,
where a and b differ.

--
pete

"Shraddha" <shraddhajosh@gmail.com> schrieb im Newsbeitrag
news:1180240585.955955.40150@j4g2000prf.googlegroups.com...

To avoid a leading bit to be interpreted as a negative value

> Why count is unsigned?

Why not? After all there would never be a negative number of bits counted.
unsigned char would have been sufficient here, don't think there are many
machine out there with more than 256 bits in an int...

> And why you XOR the inputs?

To find the bits that are set differently in the 2 operands

Bye, Jojo

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