CODEWITHSUNDEEP
c++gccfloating-point
Msegade
Msegade

Reputation: 488

Minimum (positive) floating point number (closest to zero)

I'm trying to find the minimum (positive) value (closest to zero) that I can store in a single precission floating point number. Using the <limits> header I can get the value, but if I make it much smaller, the float can still hold it and it gives the right result. Here is a test program, compiled with g++ 5.3.0.

#include <limits>
#include <iostream>
#include <math.h>

using namespace std;

int main()
{
    float a = numeric_limits<float>::max();    
    float b = numeric_limits<float>::min(); 

    a = a*2;
    b = b/pow(2,23);
    

    cout << a << endl;
    cout << b << endl;
}

As I expected, "a" gives infinity, but "b" keeps holding the good result even after dividing the minimum value by 2^23, after that it gives 0.

The value that gives numeric_limits<float>::min() is 2^(-126) which I belive is the correct answer, but why is the float on my progam holding such small numbers?

Upvotes: 7

Views: 11210

Answers (3)

xaxxon
xaxxon

Reputation: 19761

I'm trying to find the minimum value (closest to zero) that I can store in a single precission floating point number

0 is the closest value to 0 that you can store in any precision float. In fact, you can store it two ways, as there is a positive and negative 0.

edit: I edited the question to specify positive but this was the correct answer for what was originally asked.

Upvotes: -4

Pete Becker
Pete Becker

Reputation: 76305

std::numeric_limits::min for floating-point types gives the smallest non-zero value that can be represented without loss of precision. std::numeric_limits::lowest gives the smallest representable value. With IEEE representations that's a subnormal value (previously called denormalized).

Upvotes: 7

waleed
waleed

Reputation: 214

From wikipedia https://en.wikipedia.org/wiki/Single-precision_floating-point_format:

The minimum positive normal value is 2^−126 ≈ 1.18 × 10^−38 and the minimum positive (denormal) value is 2^−149 ≈ 1.4 × 10^−45.

So, for 

cout  << (float)pow(2,-149) 
      << "-->" << (float)pow(2,-150) 
      << "-->" <<  (float)pow(2,-151) << endl;

I'm getting: 

1.4013e-45-->0-->0

Upvotes: 3

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