Anonymous 7843 said:
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.
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 */
gcc -Wall -ansi -pedantic -lm -o my_sqrt my_sqrt.c
my_sqrt -45.9314
Calculated in 8 iterations
sqrt(-45.931400) = 6.7772708371i
-Charles
