CODEWITHSUNDEEP
c++gmpmpfr
sunhex
sunhex

Reputation: 53

How to extend the precision of mpreal precisely?

Consider the following code:

#include <iostream>
#include <mpreal.h>
using namespace std;
using mpfr::mpreal;

mpreal x("1.001",64);
mpreal y("1.0",64);
y*=x;
cout<<y<<endl;  //1
y.set_prec(128);
cout<<y<<endl;  //2

The outputs are

1.001

1.00100000000000000002081668171172168513

I wish that the second output is something like

1.00100000000000000000000000000000000000

In fact, I've learned that one can replace

y.set_prec(128);

to

y=mpreal(y.toString(),128);

But this kind of conversion is time consuming.

Is there any better/faster method?

Thanks!

Upvotes: 0

Views: 490

Answers (1)

Pavel Holoborodko
Pavel Holoborodko

Reputation: 557

It is better to setup precision globally, so that all mpreal variables will be created with such precision by default. Call the following function before creating any mpreal numbers:

mpfr::mpreal::set_default_prec(mpfr::digits2bits(N));

mpreal x("1.001");  // second argument is not required anymore
mpreal y("1.0");
...

where N is required precision in decimal digits.

Binary floating-point numbers don't give you actual decimal precision, it is only approximations to real numbers. More bits in representation = higher approximation accuracy. But some numbers can never be represented exactly in binary format. The 1.001 is one of them. So that you will always see non-zeros starting from some position in binary representation of 1.001 number.

Check What Every Computer Scientist Should Know About Floating Point Arithmetic for more details.

Upvotes: 1

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