Thread: usage of powl() function

Hi, this is probably a very simple problem however I cannot get a valid output from the powl() function. everytime I run it, it gives an output of 0.00. This is probably just a simple usage problem but I can't work out how to raise long double variables to the power of other long double variables. My program is an iterative calculation of pi using 4*(4arctan(1/5)-arctan(1/239). I know you could use the inbuilt arctan functions in math.h but i wanted to make it iterative. I have read the man pages on pow()/powf()/powl() and scoured google but can't get it to work.

sorry for if i'm being thick.

Code:

/*
 program works by working out pi as 4*(4arctan(1/5)-arctan(1/239)
it approximates arctan(z) as (z^1)/1-(z^3)/3+(z^5)/5-... 
*/

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

int main (int argc, const char * argv[]) {
    long double pi, arctan1, arctan2, divisor;
	bool sign;
	int count;
	sign = 0;
	count = 0;
	divisor = 1;
	printf("\n");
	while (count < 100) {
		
		if (sign == 0) 
			
		{
			arctan1 += (( powl( (1/5) , divisor))/divisor );
			arctan2 += (( powl ( (1/239) , divisor))/divisor );
			sign = 1;
			
		}
		
		else 
			
		{
			arctan1 -= (( powl ( (1/5) , divisor))/divisor );
			arctan2 -= (( powl ( (1/239) , divisor))/divisor );
			sign = 0;
			
		}
		
		
		pi=4*((4*arctan1)-arctan2);
		
		printf("\r%.050Lf", pi);
		count++;
		divisor +=2;
		
	}
	
	return 0;
	
}

Ok thanks a lot, i've changed it to using decimals instead of dividing two ints. how would i set it to divide them by using a floating point calculation in powl()?

Code:

/*
 program works by working out pi as 4*(4arctan(1/5)-arctan(1/239)
it approximates arctan(z) as (z^1)/1-(z^3)/3+(z^5)/5-... 
*/

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

int main (int argc, const char * argv[]) {
    long double pi, arctan1, arctan2, divisor;
	bool sign;
	int count;
	
	sign = 0;
	count = 0;
	divisor = 1;
	printf("\n");
	while (count < 100) {
		
		if (sign == 0) 
			
		{
			arctan1 += (( powl( (0.2) , divisor))/divisor );
			arctan2 += (( powl ( (0.0041841) , divisor))/divisor );
			sign = 1;
			
		}
		
		else 
			
		{
			arctan1 -= (( powl ( (0.2) , divisor))/divisor );
			arctan2 -= (( powl ( (0.0041841) , divisor))/divisor );
			sign = 0;
			
		}
		
		
		pi=4*((4*arctan1)-arctan2);
		
		printf("\r%.050Lf", pi);
		count++;
		divisor +=2;
		
	}
	
	return 0;
	
}

thank you very much for your help, as you've probably guessed, i'm extremely new to C programming.
ok last question (sorry)
how can I implement more precision than long double, because after 20 iterations it doesn't seem to change the value of pi. i'm guessing this is due to rounding due to it running out of usable bits in memory. (maybe i'm wrong)

i've heard about __float128 but i'm not sure if this would be tricky to implement, there's not much i can find about it.

Code:

/*
 program works by working out pi as 4*(4arctan(1/5)-arctan(1/239)
it approximates arctan(z) as (z^1)/1-(z^3)/3+(z^5)/5-... 
*/

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

int main (int argc, const char * argv[]) {
    long double pi, arctan1, arctan2, divisor, a, b;
	bool sign;
	int count;
	a = (1.0)/(5.0);
	b = (1.0)/(239.0);
	sign = 0;
	count = 0;
	divisor = 1;
	printf("\n");
	while (count < 2000) {
		
		if (sign == 0) 
			
		{
			arctan1 += (( powl ( a , divisor))/divisor );
			arctan2 += (( powl ( b , divisor))/divisor );
			sign = 1;
			
		}
		
		else 
			
		{
			arctan1 -= (( powl ( a , divisor))/divisor );
			arctan2 -= (( powl ( b , divisor))/divisor );
			sign = 0;
			
		}
		
		
		pi=4*((4*arctan1)-arctan2);
		
		printf("%.0100Lf\n", pi);
		count++;
		divisor +=2;
		
	}
	printf("\n");
	return 0;
	
}

thanks a lot for the help. i've found it extremely useful. I will continue learning C programming before I move onto another language or compiler extensions.