CODEWITHSUNDEEP
c++floating-point
Nusrat Nuriyev
Nusrat Nuriyev

Reputation: 1174

Loss of precision while working with double

Could we work with big numbers up to 10^308. How can I calculate the 11^105 using just double?

The answer of (11^105) is: 22193813979407164354224423199022080924541468040973950575246733562521125229836087036788826138225193142654907051

Is it possible to get the correct result of 11^105? As I know double can handle 10^308 which is much bigger than 11^105. I know that this code is wrong:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <iomanip>
using namespace std;

int main()
{
    double n, p, x;
    cin >> n >> p;
    //scanf("%lf %lf", &n,&p);
    x = exp(log((double)n)*p);
    //printf("%lf\n", x);
    cout  << x <<endl;
    return 0;
}

Thanks.

Upvotes: 1

Views: 168

Answers (2)

Mohit Jain
Mohit Jain

Reputation: 30489

Double can hold very large results, but not high precision. In constrast to fixed point numbers, double is floating point real number. This means, for the same accuracy double can shift the radix to handle different range of number and thus you see high range.

For your purpose, you need some home cooked big num library, or you can find one readily available and written by someone else.

BTW my home cooked recipe gives different answer for 11105 Confirmed with this haskell code

Upvotes: 2

Haozhun
Haozhun

Reputation: 6511

double usually has 11bit for exp (-1022~1023 normalized), 52bit for fact and 1bit for sign. Thus 11^105 cannot be represented accurately.

For more explanation, see IEEE 754 on Wikipedia

Upvotes: 5

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