我有一个功能非常适合不太大的值:
public BigInteger GetFrom(decimal value)
{
var decimalPlaces = 20;
var factor = (decimal) Math.Pow(10, decimalPlaces);
return new BigInteger(value * factor);
}
它正确转换:
123m = 12300000000000000000000
123.123456789m = 12312345678900000000000
0.123m = 12300000000000000000
0.123456789m = 12345678900000000000
0.123456789012345678902345678901234567890m = 12345678901234567890
1.123456789012345678902345678901234567890m = 112345678901234567890
但是如下所示:12345678901234567890.12345678901234567890m。当然因为12345678901234567890.12345678901234567890m * Math.Pow(10,20)对于小数来说太大了,但我不需要这个十进制我需要这个像BigInteger之前的例子一样
12345678901234567890.12345678901234567890m = 1234567890123456789012345678901234567890
买,我不知道什么/如何解决这个问题的最佳方法...
好吧,基本上遵循jdweng的建议:
public BigInteger GetFrom(decimal value)
{
DecimalPlaces = 20;
string strValue = HasDecimalPlaces(value) ? ConvertAValueWithDecimalPlaces(value) : ConvertARoundedValue(value);
return BigInteger.Parse(strValue);
}
private static bool HasDecimalPlaces(decimal value)
{
return ! Math.Round(value).Equals(value) || value.ToString(CultureInfo.InvariantCulture).Contains(".");
}
private string ConvertAValueWithDecimalPlaces(decimal value)
{
var commaLeftRight = value.ToString(CultureInfo.InvariantCulture).Split('.');
return commaLeftRight[0] + commaLeftRight[1].PadRight(DecimalPlaces, '0').Substring(0, DecimalPlaces);
}
private string ConvertARoundedValue(decimal value)
{
return value.ToString(CultureInfo.InvariantCulture) + new string('0', DecimalPlaces);
}