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std::numeric_limits::epsilon

From cppreference.com
 
 
 
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Type trait constants
 
 
static T epsilon();
(until C++11)
static constexpr T epsilon();
(since C++11)

Returns the machine epsilon, that is, the difference between 1.0 and the next value representable by the floating-point type T. It is only meaningful if std::numeric_limits<T>::is_integer == false.

Contents

[edit] Return value

T std::numeric_limits<T>::epsilon()
/* non-specialized */ T();
bool false
char 0
signed char 0
unsigned char 0
wchar_t 0
char16_t 0
char32_t 0
short 0
unsigned short 0
int 0
unsigned int 0
long 0
unsigned long 0
long long 0
unsigned long long 0
float FLT_EPSILON
double DBL_EPSILON
long double LDBL_EPSILON

[edit] Exceptions

(none) (until C++11)
noexcept specification:  
noexcept
  
(since C++11)

[edit] Example

Demonstrates the simplistic use of machine epsilon to compare floating-point values (this approach doesn't work for the subnormal values, for which the epsilon is meaningless)

#include <cmath>
#include <limits>
#include <iomanip>
#include <iostream>
#include <type_traits>
#include <algorithm>
 
template<class T>
typename std::enable_if<!std::numeric_limits<T>::is_integer, bool>::type
    almost_equal(T x, T y, int ulp)
{
    // the machine epsilon has to be scaled to the magnitude of the larger value
    // and multiplied by the desired precision in ULPs (units in the last place)
    return std::abs(x-y) <= std::numeric_limits<T>::epsilon() * std::abs(x+y) * ulp;
}
int main()
{
    double d1 = 0.2;
    double d2 = 1 / std::sqrt(5) / std::sqrt(5);
 
    if(d1 == d2)
            std::cout << "d1 == d2\n";
    else
            std::cout << "d1 != d2\n";
 
    if(almost_equal(d1, d2, 2))
            std::cout << "d1 almost equals d2\n";
    else
            std::cout << "d1 does not almost equal d2\n";
}

Output:

d1 != d2
d1 almost equals d2

[edit] See also

(C++11)(C++11)
next representable floating point value towards the given value
(function) [edit]