std::numeric_limits<T>::epsilon
来自cppreference.com
< cpp | types | numeric limits
static T epsilon() throw(); |
(C++11 前) | |
static constexpr T epsilon() noexcept; |
(C++11 起) | |
返回机器 epsilon ,即 1.0 与浮点类型 T
的下个可表示值的差。它仅若 std::numeric_limits<T>::is_integer == false 才有意义。
返回值
T
|
std::numeric_limits<T>::epsilon() |
/* non-specialized */ | T()
|
bool | false |
char | 0 |
signed char | 0 |
unsigned char | 0 |
wchar_t | 0 |
char8_t (C++20) | 0 |
char16_t (C++11) | 0 |
char32_t (C++11) | 0 |
short | 0 |
unsigned short | 0 |
int | 0 |
unsigned int | 0 |
long | 0 |
unsigned long | 0 |
long long (C++11) | 0 |
unsigned long long(C++11) | 0 |
float | FLT_EPSILON |
double | DBL_EPSILON |
long double | LDBL_EPSILON |
示例
演示用机器 epsilon 比较浮点值是否相等
运行此代码
#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 values used // 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 // unless the result is subnormal || std::abs(x-y) < std::numeric_limits<T>::min(); } int main() { double d1 = 0.2; double d2 = 1 / std::sqrt(5) / std::sqrt(5); std::cout << std::fixed << std::setprecision(20) << "d1=" << d1 << "\nd2=" << d2 << '\n'; 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"; }
输出:
d1=0.20000000000000001110 d2=0.19999999999999998335 d1 != d2 d1 almost equals d2
参阅
(C++11)(C++11) (C++11)(C++11)(C++11)(C++11) |
趋向给定值的下个可表示浮点值 (函数) |