C++ Adding 1 to very small number?

问题

I'm just trying to compute a good sigmoid function in C++ (and efficient). So i have to do something like:
1/(1 + exp(-x))

The problem is, when X becomes big (or even small), the result of 1 + e turns to be 0 or 1

For example,
1 + exp(-30) = 1
But this is incorrect...

How can we add very small (or big) numbers easily and efficiently ?

Datatype I am using : double

Here is the code snippet:

double Quaternion::sigmoidReal(double v){
    return 1.0 / ( 1.0 + exp(-v) ) ; 
}

Thanks !

回答1:

I think you need to set the precision of cout:

#include <iostream>
#include <iomanip>

int main()
{
    std::cout << std::fixed;
    std::cout << std::setprecision(30);
    std::cout << (1 + exp(-30)) << std::endl;
    getchar();
    return 0;
}

Output:

1.000000000000093480778673438181

Note: Like people have noted in the comments, part of the output is noise (which at the time, I didn't consider). I was mostly just trying to demonstrate that you can put an arbitrary number into setprecision.

This has to do with the maximum amount of information that can be saved in 1 double. Part of the bits is used for the exponent, and part for the actual digits. A more accurate way would be to print the number like this (without setting the precision):

std::cout << 1 << "+" << exp(-30) << std::endl;

Output:

1+9.35762e-14

Which still leaves the problem of saving a double with that actual value. However, it is possible, and wikipedia has a list of libraries that help with this: Link. The related article also has some more explanation.

来源：https://stackoverflow.com/questions/36547763/c-adding-1-to-very-small-number

标签

c++

floating-point

double

precision