Question

i have a problem in calc of pi, when i put long double in variable appears a big number, but not the correct value, if you change for calcf function with double it works.
My code:

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

long double calc(int n, double denominador) {
    //printf("%lf e",(pow(-1,n))/denominador);
    //printf("%d\n",(denominador));
    long double var = (long double) ((4.0*(pow(-1.0,n)))/denominador);
    printf("%Lf\n",(long double)var);
    return var;
}

double calcf(int n, int denominador) {
    //printf("%lf e",(pow(-1,n))/denominador);
    //printf("%d\n",(denominador));
    double var = (4.0*(pow(-1.0,n)))/denominador;
    //printf("%lf",var);
    return var;
}

int main() {
    int NUMERO = 100000000;
    long double pi = 4L;
    double pif = 4;
    int i;
    int n=1;
    printf("%d e %d",sizeof(double),sizeof(long double));
    for (i=3;i<NUMERO;i+=2) {
            pi += calc(n,(double) i);
        //pif += calcf(n,i);

            n++;
    }
    printf("PI: %1.50Lf\n",pi);
}

What i’m doing wrong?

Thank you.

Answer 1

Well, this program has many issues; but the worst one is that you’re using Gregory’s series, which is just about the slowest algorithm out there. You’re not even going to get double precision with that, let alone long double. You would have to let it run through about 10¹⁶ steps just for double.

Mathematicians use −1ⁿ to indicate an alternating sign, which is fine in that context, but using pow() to calculate it is just silly, especially since it’s a floating point operation. Instead, use an integer that starts at -1 and keep doing i = -i.

You need to be more careful with your types and constants. 4L is a long integer, not a long double. You want 4.0L. Of course the compiler casts it anyway, but it’s bad practice. You’re using plain old double constants like 4.0 when you’re aiming for long double precision.

It’s also silly to keep multiplying everything by 4.0 at every step. You can just do it at the end.