Inverting the exponent of a long double gives me an insane result

Question

I am trying to invert the exponent of a long double.

Suppose x = 3.5e1356. I want x to be 3.5e-1356.

I have this code:

long double x = 3.5e1356L;
int exponent;
long double fraction = frexpl(x, &exponent);

// recreate number with an inverted exponent
long double newNumber =  ldexpl(fraction, -exponent);

after this code, newNumber is 1.14732677619641872902e-1357

that has nothing to do with the original number.

What am I missing?

Answer 1

Use the fact that

a * (10^n) * (10^m) = a * 10^(n+m) 

If you calculate m so that it is -2n you'll get:

a * (10^n) * (10^-2n) = a * 10^(n+(-2n)) =  a * 10^-n

In other words - just multiply the original number with 10^-2n

I would try something like this:

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

int main(void) {
    long double x = 8.9e-100;
    long double y = 0.0;
    int exp10;

    if (x != 0.0)
    {
        exp10 = log10l(fabsl(x));
        exp10 = (exp10 >= 0) ? -2*exp10 : -2*(exp10-1);
        y = x * powl(10, exp10);
    }

    printf("%.36Le\n", x);
    printf("%.36Le\n", y);

    return 0;
}

Example for 3.5 * 10^12 :

3.5 * 10^12 * 10^-24 --> 3.5 * 10^-12