1.#INF00 problem while trying to calculate pi

[valenwalen]

I did my program to calculate pi but for some reason, every single time I run the program, screen says "1.#INF00":

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

int main (void)
{
    float a;
    int i, x, b, c, d, e, f;
    x=3;
    a=1.0;
    b=1;
    c=4;
    d=5;
    e=1;
    f=3;
    for (i=1; i<=9999; i=i+1)
    {
        a=a*(x+((e*b)/(c*d*f)));
        x=x+5;
        b=b+2;
        c=c+3;
        d=d+3;
        e=e+1;
    }
    printf("%f", a);
    return 0;
}

[weaknessforcats]

1.#INF00 is an IEEE floating point exception. There are several kinds of these. This one is a positive infinity and it means your float variable is not in floating point format.

That suggests your loop doesn't work.

I suggest you set a breakpoint at the start of your loop and step the loop to see if the variable a is still in floating point format. If you see a decimal number in your debugger, then a is still in float format. Otherwise, you will see the 1.#INF00.

[Banfa]

I'm not sure where you got the formula from but it is very wrong

Consider ((e*b)/(c*d*f))

All variables are integers so arithmetic is integer based and they have the following characteristics

Initial
Variable Value Increment
b 1 2
c 4 3
d 5 3
e 1 1
f 3 0

Then for these values
c > b is always true
d > e is always true

Therefore (e*b) < (c*d*f) is always true* and since integer division is in operation ((e*b)/(c*d*f)) is always 0.


* Actually this isn't quite true because towards the end of the iteration (c*d*f) actually wraps (gets bigger than an integer can hold) and on my machine this results in a negative number but the absolute value of this number is always bigger than (e*b) but that is a fluke.

Since ((e*b)/(c*d*f)) is always 0 your calculation becaomes

    x=3;
    a=1.0;
    
    for (i=1; i<=9999; i=i+1)
    {
        a=a*x;
        x=x+5;
    }

x increments by 5 every time it is getting larger all the time a is getting exponentially larger all the time. It calculates

3 * 8 * 13 * 18 * ... * 49983 * 49988 * 49993 * 49998

This is neither PI nor does it fit into a float.


In fact a takes the value 1.#INF on the 130th iteration of the loop, on the 129th iteration it had the value 7.01924e306, at this point x had a value of 648.

[valenwalen]

How could I change my program in order to obtain an aproximation to pi? because I need to do 10 similar programs and I got most of the ideas from the webpage http://mathworld.wolfram.com/PiFormulas.html, but so far I've only been able to program Leibniz's

[Banfa]

Well you need to make sure you get the logic of your implementation correct. It is not clear to me which of those 130+ formulas you where trying to implement.

[valenwalen]

You are absolutely right, my apologies. I was trying to do Beeler's, Sloane's and Wallis'. The program I previously posted was an attempt of Beeler's.