How to get correct floating-point precision in Qt and C++?

Question

I am studying Qt, I find it really amazing and a bit easier because I've already used MFC in Microsoft.

I've developed a scientific calculator for educational use. But what faced me is:

I want float high precision like in real-world calculators so after searching I've found this function:

    long double ldRes =  std::sqrt(3);
    QMessageBox m;
    m.settext(QString::number(ldRes, 'g', 30));
    m.exec();

The output:

1.73205080756887719317660412344

While the output, according the The Casio Calculator, of the same operation to the same precision, should be:

1.73205080756887729352744634151

So it is apparent that the result is not the same so can anyone advise me how to get correct float-precision in my applications using Qt? or any other library?

Should I use GMP MPIR MPFR?

Thank you!

How much precision do you want? 10 decimal digits? 20? Arbitrary? — ComicSansMS, Mar 23 '20 at 22:00
The problem has nothing to do with Qt but with the floating point (IEEE 754 standard), there are many libraries that handle the implementation of these operations without losing precision since they do not implement the standard that you could use. — eyllanesc, Mar 23 '20 at 22:00
DUPE OF: https://stackoverflow.com/questions/588004/is-floating-point-math-broken — selbie, Mar 23 '20 at 22:03
If this is Visual Studio I think `long double` is the same as `double` — drescherjm, Mar 23 '20 at 22:09
@AsteroidsWithWings: For example the casio calculator: https://keisan.casio.com/calculator — Maestro, Mar 23 '20 at 22:09
Yes to produce an application like that calculator, you may want to use a large math library like GMP or Boost::Multiprecision — drescherjm, Mar 23 '20 at 22:13
@drescherjm I hear ya, but since the result starts to deviate at ~15-17 decimal significant figures, either the implementation's `sqrt` impl is badly internally broken, or the types differ on the OP's platform — Asteroids With Wings, Mar 23 '20 at 22:18

score 5 · Answer 1 · answered Mar 23 '20 at 21:59

5

When you use std::sqrt and pass a double (well, an int, which will end up being converted to double), the result is a double.

Converting that result to long double achieves nothing but gaining a ton of low-value bits with no relation to the result of the square-root operation!

I don't know what "other calculators" you're using, but something about the difference between the two approaches leads to those different results.

std::sqrtl works with long doubles, or you can pass a long double argument to std::sqrt which picks the right overload.

// Effectively equivalent:
long double a = std::sqrtl(3);
long double b = std::sqrtl(3);
long double c = std::sqrt(3.0L);

Now, your calculation has the precision of a long double from end to end, and ought to be mathematically correct to that level of precision.

answered Mar 23 '20 at 21:59

Asteroids With Wings

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Converting a `double` to a `long double` does not “[gain] a ton of low-value bits” except for zero bits. A conversion to a more precise type does not change the value. – Eric Postpischil Mar 26 '20 at 10:44
@EricPostpischil What? You said it yourself: zero bits (by "low-value" I'm referring to their low magnitude, the fact that they are on the least-significant side of the overall value). And those zero bits are _wrong_. They make the result mathematically incorrect. No, the value isn't different from what it was as a `double`, but it _is_ different from what it will be in any tool that gives that mathematical result to that precision correctly. And that's what I said: _"achieves nothing but gaining a ton of low-value bits **with no relation to the result of the square-root operation**"_ – Asteroids With Wings Mar 26 '20 at 11:07
The language in the answer should be more precise and explicit. As humans use language, “digits” or “bits” may refer to either to significant non-zero digits or bits or to formal bits in a representation. This is particularly so in the case of floating-point discussion, where people sometimes refer to unexpected or incompletely understood values has having various noise or “randomness” in the low bits. Even if you are using language in a specific deliberate way, readers cannot know that unless they are told or at least there are indications of it. – Eric Postpischil Mar 26 '20 at 11:33
@EricPostpischil Feel free to write your own. I stand by my wording - nobody else seems to have misinterpreted it. But I do appreciate your feedback. Take care! – Asteroids With Wings Mar 26 '20 at 15:19

How to get correct floating-point precision in Qt and C++?

1 Answers1