C++ 如何简化这个模板，使其不专门化？

Question

我试图建立一个模板来比较基于 https://floating-point-gui.de/errors/comparison/ 的浮点数或 double 数对于 C++。该版本要求将 epsilon 值传递给代码，但我想为系统使用默认的 FLT_EPSILON、FLT_MIN、FLT_MAX、DBL_EPSILON、DBL_MIN 和 DBL_MAX，并且在 float.h 中可用。

有没有办法在如下语句中使用 typename:

T epsilon;
if ( T == DBL_TYPE ) epsilon = DBL_EPSILON
else epsilon = FLT_EPSILON  ????

这将使例程成为例程，因此我不必专门化它，这会导致单一定义规则 (ODR) 问题。

我是模板的新手，所以我不知道如何提问。

template <typename T>
bool IsNearEqual( T test_1, T test_2 )
{
    if ( test_1 == test_2 ) return true;
    return false;
}

template <>
bool IsNearEqual( float test_1, float test_2 )
{
    const float FLT_NORMAL = ( (long)1 << 23 ) * FLT_EPSILON;
    if ( test_1 == test_2 )
    { // shortcut, handles infinities
        return true;
    }
    float test_1Abs = dfAbs( test_1 );
    float test_2Abs = dfAbs( test_2 );
    float test_1_2AbsDif = dfAbs( test_1 - test_2 );
    float test_1_2Sum = test_1Abs + test_2Abs;

    if ( test_1 == 0 || test_2 == 0 || test_1_2Sum < FLT_NORMAL )
    { // test_1 or test_2 is zero or both are extremely close to it relative error is less meaningful here
        return test_1_2AbsDif < ( FLT_EPSILON * FLT_NORMAL );
    }
    else
    { // use relative error
        return test_1_2AbsDif / dfMin( test_1_2Sum, FLT_MIN ) < FLT_EPSILON;
    }
    return false;
}
template <>
bool IsNearEqual( double test_1, double test_2 )
{
    const double DBL_NORMAL = ( (long long)1 << 52 ) * DBL_EPSILON;

    if ( test_1 == test_2 )
    { // shortcut, handles infinities
        return true;
    }
    double test_1Abs = dfAbs( test_1 );
    double test_2Abs = dfAbs( test_2 );
    double test_1_2AbsDif = dfAbs( test_1 - test_2 );
    double test_1_2Sum = test_1Abs + test_2Abs;

    if ( test_1 == 0 || test_2 == 0 || test_1_2Sum < DBL_NORMAL )
    { // test_1 or test_2 is zero or both are extremely close to it relative error is less meaningful here
        return test_1_2AbsDif < ( DBL_EPSILON * DBL_NORMAL );
    }
    else
    { // use relative error
        return test_1_2AbsDif / dfMin( test_1_2Sum, DBL_MAX ) < DBL_EPSILON;
    }
    return false;
}

template <typename T, typename U>
T dfMin( T test_1, U test_2 )
{
    T out;
    out = test_1;
    if ( out > test_2 ) out = (T)test_2;

    return out;
}

template <typename T, typename U>
T dfMax( T test_1, U test_2 )
{
    T out;
    out = test_1;
    if ( test_2 > out ) out = (T)test_2;

    return out;
}

template <typename T>
T dfAbs( T test )
{
    T out;
    out = test;
    if ( test < 0.0 ) out = -test;

    return out;
}

谢谢，
罗纳克

Answer 1

这是带有一点 SFINAE 和使用 std::numeric_limits 的代码正如评论中所建议的那样。

template <class T, std::enable_if_t<std::is_floating_point<T>{}>* = nullptr>
bool IsNearEqual(T test_1, T test_2) {
  using nl = std::numeric_limits<T>;

  const T NORMAL = ((long)1 << nl::digits) * nl::epsilon();
  if (test_1 == test_2) {  // shortcut, handles infinities
    return true;
  }
  float test_1Abs = dfAbs(test_1);
  float test_2Abs = dfAbs(test_2);
  float test_1_2AbsDif = dfAbs(test_1 - test_2);
  float test_1_2Sum = test_1Abs + test_2Abs;

  if (test_1 == 0 || test_2 == 0 ||
      test_1_2Sum < NORMAL) {  // test_1 or test_2 is zero or both are extremely
                               // close to it relative error is less meaningful
                               // here
    return test_1_2AbsDif < (nl::epsilon() * NORMAL);
  } else {  // use relative error
    return test_1_2AbsDif / dfMin(test_1_2Sum, nl::max()) < nl::epsilon();
  }
  return false;
}