function without arithematic operator

hello,
Is it possible to write a math library function sqrt in C without
arithmatic operators(*,+)?
can anyone give me hints to do that?
regards,
rahul

rahul8143@gmail.com writes:
[color=blue]
> Is it possible to write a math library function sqrt in C without
> arithmatic operators(*,+)?[/color]

There is no good reason to do so.
--
"If I've told you once, I've told you LLONG_MAX times not to
exaggerate."
--Jack Klein

rahul8143@gmail.com wrote:[color=blue]
>
> hello,
> Is it possible to write a math library function sqrt in C without
> arithmatic operators(*,+)?
> can anyone give me hints to do that?[/color]

Use division and subtraction instead. :-)

Of course, there are only two reasons to do so that I can think of:

(1) A nonsensical homework assignment.
(2) A troll.

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

In article <1122991677.514736.32910@z14g2000cwz.googlegroups. com>,
<rahul8143@gmail.com> wrote:[color=blue]
>
>
> hello,
> Is it possible to write a math library function sqrt in C without
> arithmatic operators(*,+)?
> can anyone give me hints to do that?
> regards,
> rahul[/color]

If it's in the math library it doesn't have to be implemented
in C, so if the processor has a sqrt instruction you can just
use that. That's cheating (kind of) because it's likely that the
processor itself will do operations similar to * and + in order
to execute.

exp(log(x)/2.0) is workable, but divide is an arithmetic
operator, and although you didn't specifically mention it in
your list and I can't tell from your words whether the list
is meant to be exhaustive or to exemplify. Again, exp and
log are likely to do some arithmetic operations behind
the scenes.

It's _probably_ possible to use shifts and bitwise operations
to implement an integer sqrt function. Another integer-only
solution would be something really lame like:

int sqrt(int x)
{
if (x < 4) return 1;
if (x < 9) return 2;
if (x < 16) return 3;
if (x < 25) return 4;
if (x < 36) return 5;
/* and so on, up to sqrt(INT_MAX) */
}

But perhaps you have handed in your homework by now and nothing
I have said is helpful.
--
7842++

"Anonymous 7843" <anon7843@example.com> wrote in message
news:uhPHe.27079$HV1.5748@fed1read07...[color=blue]
> In article <1122991677.514736.32910@z14g2000cwz.googlegroups. com>,
> <rahul8143@gmail.com> wrote:[color=green]
> >
> >
> > hello,
> > Is it possible to write a math library function sqrt in C without
> > arithmatic operators(*,+)?
> > can anyone give me hints to do that?
> > regards,
> > rahul[/color]
>
> If it's in the math library it doesn't have to be implemented
> in C, so if the processor has a sqrt instruction you can just
> use that. That's cheating (kind of) because it's likely that the
> processor itself will do operations similar to * and + in order
> to execute.
>
> exp(log(x)/2.0) is workable, but divide is an arithmetic
> operator, and although you didn't specifically mention it in
> your list and I can't tell from your words whether the list
> is meant to be exhaustive or to exemplify. Again, exp and
> log are likely to do some arithmetic operations behind
> the scenes.
>
> It's _probably_ possible to use shifts and bitwise operations
> to implement an integer sqrt function. Another integer-only
> solution would be something really lame like:
>
> int sqrt(int x)
> {
> if (x < 4) return 1;
> if (x < 9) return 2;
> if (x < 16) return 3;
> if (x < 25) return 4;
> if (x < 36) return 5;
> /* and so on, up to sqrt(INT_MAX) */
> }
>
> But perhaps you have handed in your homework by now and nothing
> I have said is helpful.[/color]

The code below uses an iterative algorithm to calculate the sqrt of a
double, without any addition or multiplication operations (although
subtraction and division are both used). Comments are welcome.

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

#define FLIP_SGN(x) -x

double my_abs(double x) {
if(x<0)
return FLIP_SGN(x);
else
return x;
} /* my_abs */

double my_sqrt(double x) {
double ans=0, guess=x, error=1e-12, error2;
int count=0;
error2 = my_abs(ans-guess);

while(error2>error){
ans = (guess-x/FLIP_SGN(guess)) / 2.0; /* Newton's method approximation
*/
error2 = my_abs(ans-guess);
guess = ans;
count--;
} /* while */
printf("Calculated in %d iterations\n", FLIP_SGN(count));

return ans;
} /* my_sqrt */

int main(int argc, char **argv) {
int imag=0;
char *str_test;
double x, ans;

if(argc>1) {
x = strtod(argv[1], &str_test);
if(argv[1]!=str_test) {
if(x<0) {
imag = 1;
x = FLIP_SGN(x);
} /* if */
ans = my_sqrt(x);
if(imag)
printf("sqrt(%f) = %.10fi\n", FLIP_SGN(x), ans);
else
printf("sqrt(%f) = %.10f\n", x, ans);
} /* if */
else
printf("ERROR: Improper input provided; exiting!\n");
} /* if */
else
printf("ERROR: No input provided; exiting!\n");

return 0;
} /* main */
[color=blue]
>gcc -Wall -ansi -pedantic -lm -o my_sqrt my_sqrt.c
>my_sqrt -45.9314[/color]
Calculated in 8 iterations
sqrt(-45.931400) = 6.7772708371i

-Charles

On 2 Aug 2005 07:07:57 -0700, rahul8143@gmail.com wrote:
[color=blue]
>hello,
>Is it possible to write a math library function sqrt in C without
>arithmatic operators(*,+)?
>can anyone give me hints to do that?
>regards,
>rahul[/color]
exp(log(n)/2);

is probably one of the least efficient ways.

--
Al Balmer
Balmer Consulting
removebalmerconsultingthis@att.net

Kenneth Brody wrote:[color=blue]
>
> rahul8143@gmail.com wrote:[color=green]
> >
> > hello,
> > Is it possible to write a math library function sqrt in C without
> > arithmatic operators(*,+)?
> > can anyone give me hints to do that?[/color]
>
> Use division and subtraction instead. :-)
>
> Of course, there are only two reasons to do so that I can think of:
>
> (1) A nonsensical homework assignment.
> (2) A troll.[/color]

#include <errno.h>
#include <math.h>

double sqrt(double x)
{
if (x > 0) {
const double a = x;
double b = a / 2 - -0.5;

do {
x = b;
b = (a / x - -x) / 2;
} while (x > b);
} else {
if (0 > x) {
errno = EDOM;
x = -HUGE_VAL;
}
}
return x;
}

--
pete

You can also use SQR function to do that.(If SQR is not an arithmetic
operator)
while not (answer)
{
i++;
if(i>number){
answer=true;
}
}

you can also calculate the decimals with a small improvement.