在採用C++編寫算法時,經常需要判斷兩個浮點數是否相等。由於計算精度的原因,採用“==”運算符是不可行的。下面給出採用C++11標準判斷兩個浮點數是否相等的代碼:
// Test whether two float or double numbers are equal.
// ulp: units in the last place.
template <typename T>
typename std::enable_if<!std::numeric_limits<T>::is_integer, bool>::type
IsAlmostEqual(T x, T y, int ulp = 2) {
// 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::fabs(x - y) <
std::numeric_limits<T>::epsilon() * std::fabs(x + y) * ulp
// unless the result is subnormal
|| std::fabs(x - y) < std::numeric_limits<T>::min();
}
上述代碼中,std::enable_if<!std::numeric_limits<T>::is_integer, bool>::type
的含義是:如果數據類型T
不是整數類型,則返回值類型爲bool
,否則會導致編譯錯誤。也就是說,只允許進行非整數數值類型的比較。當然這裏有個隱含前提,T
必須是數值類型,不可能拿一個類似於Person
的類型去實例化該函數,如果這樣做,則std::numeric_limits<T>
肯定通不過,會報編譯錯誤。
下面是示例代碼:
#include <cmath>
#include <limits>
#include <iostream>
#include <type_traits>
// Test whether two float or double numbers are equal.
// ulp: units in the last place.
template <typename T>
typename std::enable_if<!std::numeric_limits<T>::is_integer, bool>::type
IsAlmostEqual(T x, T y, int ulp = 2)
{
// 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::fabs(x - y) < std::numeric_limits<T>::epsilon() * std::fabs(x + y) * ulp
// unless the result is subnormal
|| std::fabs(x - y) < std::numeric_limits<T>::min();
}
int main()
{
double d1 = 0.2;
double d2 = 1 / std::sqrt(5) / std::sqrt(5);
std::cout << "d1 = " << d1 << std::endl;
std::cout << "d2 = " << d2 << std::endl;
if (d1 == d2)
{
std::cout << "d1 == d2" << std::endl;
}
else
{
std::cout << "d1 != d2" << std::endl;
}
if (IsAlmostEqual(d1, d2))
{
std::cout << "d1 almost equals d2" << std::endl;
}
else
{
std::cout << "d1 does not almost equal d2" << std::endl;
}
return 0;
}
- 1
Linux系統下,採用GCC編譯器的編譯命令爲:
g++ -g -Wall -std=c++11 *.cpp -o test
輸出結果如下:
d1 = 0.2
d2 = 0.2
d1 != d2
d1 almost equals d2