我在 VB 中有这样的代码行:
Dim Sqrt As Double
Sqrt = Radius ^ 2 - (CenterX - X) ^ 2
上面语句中的参数传递的值如下:
X= -7.3725025845036161 Double
CenterX =0.0 Double
Radius= 8.0 Double
在执行上面的语句时,Sqrt
的值如下:
Sqrt 9.646205641487505 Double
现在我使用 Math
类编写了类似的 C# 逻辑:
double Sqrt = 0;
Sqrt = Math.Pow(Radius, 2) - Math.Pow((CenterX - X), 2);
使用相同的一组值,C#
代码的输出为:
Sqrt 9.6462056414874979 double
由于 C# 代码中的这一单一更改,我需要帮助,我的所有值都受到影响。我能做些什么来获得与 *VB*
源代码相似的值吗?
最佳答案
有一个difference in the precision between the VB6 and the .NET double type .两者都是 IEEE 64 位 double 类型,但 .NET CLR 在内部使用 80 位扩展精度,即您的计算在 .NET 中会更准确。
如果您必须向后兼容 VB6 精度,您可以强制 FPU(浮点单元)使用(精度较低的)64 位值。这可以使用 native _controlfp_s
来实现功能。
下面是一个代码片段,您可以使用它暂时“降级”浮点精度以实现向后兼容性。你可以像这样使用它:
用法
// default floating point precision
using (new FloatingPoint64BitPrecision())
{
// floating-point precision is set to 64 bit
}
// floating-point precision is reset to default
代码片段
/// <summary>
/// This class changes floating-point precision to 64 bit
/// </summary>
internal class FloatingPoint64BitPrecision : IDisposable
{
private readonly bool _resetRequired;
public FloatingPoint64BitPrecision()
{
int fpFlags;
var errno = SafeNativeMethods._controlfp_s(out fpFlags, 0, 0);
if (errno != 0)
{
throw new Win32Exception(
errno, "Unable to retrieve floating-point control flag.");
}
if ((fpFlags & SafeNativeMethods._MCW_PC) != SafeNativeMethods._PC_64)
{
Trace.WriteLine("Change floating-point precision to 64 bit");
errno = SafeNativeMethods._controlfp_s(
out fpFlags, SafeNativeMethods._PC_64, SafeNativeMethods._MCW_PC);
if (errno != 0)
{
throw new Win32Exception(
errno, "Unable to change floating-point precision to 64 bit.");
}
_resetRequired = true;
}
}
public void Dispose()
{
if (_resetRequired)
{
Trace.WriteLine("Resetting floating-point precision to default");
SafeNativeMethods._fpreset();
}
}
}
internal static class SafeNativeMethods
{
[DllImport("msvcr120.dll")]
public static extern void _fpreset();
[DllImport("msvcr120.dll", CallingConvention = CallingConvention.Cdecl)]
public static extern int _controlfp_s(
out int currentControl, int newControl, int mask);
public static int _CW_DEFAULT =
(_RC_NEAR | _PC_53 | _EM_INVALID | _EM_ZERODIVIDE | _EM_OVERFLOW
| _EM_UNDERFLOW | _EM_INEXACT | _EM_DENORMAL);
public const int _MCW_EM = 0x0008001f; // interrupt Exception Masks
public const int _EM_INEXACT = 0x00000001; // inexact (precision)
public const int _EM_UNDERFLOW = 0x00000002; // underflow
public const int _EM_OVERFLOW = 0x00000004; // overflow
public const int _EM_ZERODIVIDE = 0x00000008; // zero divide
public const int _EM_INVALID = 0x00000010; // invalid
public const int _EM_DENORMAL = 0x00080000; // denormal exception mask
// (_control87 only)
public const int _MCW_RC = 0x00000300; // Rounding Control
public const int _RC_NEAR = 0x00000000; // near
public const int _RC_DOWN = 0x00000100; // down
public const int _RC_UP = 0x00000200; // up
public const int _RC_CHOP = 0x00000300; // chop
public const int _MCW_PC = 0x00030000; // Precision Control
public const int _PC_64 = 0x00000000; // 64 bits
public const int _PC_53 = 0x00010000; // 53 bits
public const int _PC_24 = 0x00020000; // 24 bits
public const int _MCW_IC = 0x00040000; // Infinity Control
public const int _IC_AFFINE = 0x00040000; // affine
public const int _IC_PROJECTIVE = 0x00000000; // projective
}
关于c# - 此 VB6 操作产生问题的等效 C# 语句,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39740935/