Why do I get so precise floating-point number from std::cout?

Question

The program

int main ()
{
    long long ll = LLONG_MAX;
    float f = ll;
    std::cout << ll << '\n';
    std::cout << std::fixed << f << '\n';
    return 0;
}

gives:

9223372036854775807
9223372036854775808.000000

How is it possible? If 23-bit mantissa can have only 8,388,607 maximum value, why does cout output a 64-bit number?

If you'll notice, you don't get the same value for both values, so it isn't precise, it's an approximation — NathanOliver, Sep 11 '20 at 18:41
@NathanOliver, relatively to what I expect, it's precise, but yes, you're right — Oleksa, Sep 11 '20 at 18:46
[What Every Computer Scientist Should Know About Floating-Point Arithmetic](https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) — Jesper Juhl, Sep 11 '20 at 18:47
Related: [Is floating point math broken?](https://stackoverflow.com/questions/588004/) — Remy Lebeau, Sep 11 '20 at 19:01
Does this answer your question? [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — rsjaffe, Sep 11 '20 at 19:04

score 2 · Accepted Answer · 2020-09-11T18:56:28.537

2

You stored 2^63-1 in a float, which was rounded to 2^63 = 9223372036854775808. The powers of 2 are exactly representable.

The nearest number which is exactly representable is 2^63 + 2^40 = 9223373136366403584.

edited Sep 11 '20 at 18:56

answered Sep 11 '20 at 18:48

Can you explain a bit why + 2^40 gives the next representable number? – Oleksa Sep 11 '20 at 19:08
@olekstolar: 63 - 23 = 40. – Sep 11 '20 at 19:18
It doesn't explain for me. If exponent bits have a value of 63 and a mantissa has 1, how can the next representable be something else than 2*2^63 or 1*2^64? – Oleksa Sep 11 '20 at 19:45
@olekstolar: the decimal point is on the left. Check the definition of the mantissa. – Sep 11 '20 at 19:47

score 1 · Answer 2 · answered Sep 11 '20 at 18:49

1

long long for you is a 64 bit data type so that means LLONG_MAX has a value of 2^63 - 1. You are right in that this can't be stored in a float which only has 23 bits of mantissa, but 2^63, which is one more than LLONG_MAX is easily stored in a float. It stores 2 in the mantissa and 63 in the exponent and there you have it.

answered Sep 11 '20 at 18:49

NathanOliver

171,901
28
288
402

3

You probably mean 1.0000... in the mantissa, actually stored as .00000... – Sep 11 '20 at 18:52
1

@YvesDaoust I'm ignoring how it's actually encoded. I'm just demonstrating that it isn't really anything to store a `2^63` in a float exactly. – NathanOliver Sep 11 '20 at 18:56

Why do I get so precise floating-point number from std::cout?

2 Answers2