Adding two doubles in C++

Question

I have a problem with summing two doubles

double a = 1e+020;
double b = -4000;
double res = a + b; //result = 1e+020

Why res is equal to a?

because the -4000 become insignificant for scientific notation. — Raildex, Oct 14 '20 at 08:39
Because floating point numbers store only limited number of leading digits — fas, Oct 14 '20 at 08:39
`1e+020` is a very large number, and `4000` is a very small number. Subtracting `4000` from `1e+020` will make almost no difference. — Some programmer dude, Oct 14 '20 at 08:39
Consider that on most systems, you only have 64 bits to work with here and numbers that go as high as ~1e308 as well as including non-integers. A 64-bit integer type only goes up to ~10^19. — chris, Oct 14 '20 at 08:42
@chris the issue here is not magnitude but precision (refering to 1e308 is misleading) — 463035818_is_not_an_ai, Oct 14 '20 at 08:49
@idclev463035818, My comment is about precision. There's a huge range to cover (which is what I was trying to demonstrate) and only 64 bits to do it. — chris, Oct 14 '20 at 08:56
@chris but range and precision are not directly related because seperate bits are used for the exponent — 463035818_is_not_an_ai, Oct 14 '20 at 08:57
@idclev463035818, I think that's beside the point I was trying to make. It's confusing to a lot of people why not every integer is representable and I was trying to give some high-level intuition for _why_ so many numbers are unrepresentable. Not so much _which_ numbers, but the idea that you can pick whatever bit representation you want and it won't come anywhere close to having everything. — chris, Oct 14 '20 at 09:04

score 1 · Accepted Answer · answered Oct 14 '20 at 08:44

double has not infinite number of significant digits. You can query number of digits in base 10 with std::numeric_limits:

#include <iostream>
#include <limits>

int main(){
    std::cout << std::numeric_limits<double>::digits10;
}

Typical output is

but you are requesting 16 significant digits.

1 Answers1