将计算出的双倍数转换为十进制会纠正精度错误吗？

Question

double threshold = 1000.0;
double Mass = 0.000814;
double PartMass = 0.814;
double IncorrectPPM = Mass/PartMass * 1000000;
double CorrectedPPM = (double)((decimal)IncorrectPPM);

Console.WriteLine("Mass = {0:R}", Mass);
Console.WriteLine("PartMass = {0:R}", PartMass);
Console.WriteLine("IncorrectPPM = {0:R}", IncorrectPPM);
Console.WriteLine("CorrectedPPM = {0:R}", CorrectedPPM);
Console.WriteLine("Is IncorrectPPM over threshold? " + (IncorrectPPM > threshold) );
Console.WriteLine("Is CorrectedPPM over threshold? " + (CorrectedPPM > threshold) );

Mass = 0.000814
PartMass = 0.814
IncorrectPPM = 1000.0000000000002
CorrectedPPM = 1000
Is IncorrectPPM over threshold? True
Is CorrectedPPM over threshold? False

Answer 1

有人知道计算机是如何知道在这种情况下，它应该在转换为十进制时将1000.00000000002值更改为1000的吗？

首先是演员阵容：

(decimal)IncorrectPPM

等效于构造函数调用，请参见此处的So：

new decimal(IncorrectPPM)

那意味着

1000.0000000000002 
               ^ ^  
            15th 17th significant digit

将四舍五入为：

1000.00000000000 
               ^ 
            15th significant digit

我可以依靠这个技巧来避免重复计算的精度问题吗？

1000.000000000006
               ^  
            15th significant digit

1000.00000000001
               ^  
            15th significant digit

double threshold = 1000.0001;

将incorrectppm的强制转换从：

double CorrectedPPM = (double)((decimal)IncorrectPPM);

致：

/* 1000.000000000006 will be rounded to 1000.0000 */
double CorrectedPPM = Math.Round(IncorrectPPM, 4);

使用math.round()函数，但要小心math.round()表示小数非有效数字