c# - C#中的浮点数到十进制转换

Question

int i = 1;        // -2,147,483,648 to 2,147,483,647
float f = 2.1f;      // -3.402823e38 to 3.402823e38
long l = 3;       // -922337203685477508 to 922337203685477507
double dbl = 4.5;   // -1.79769313486232e308 to 1.79769313486232e308
decimal dec = 5.2m;  // -79228162514264337593543950335 to 79228162514264337593543950335

dec = i;     // No error
dec = l;     // No error

dec = f;    // Compiler error
dec = dbl;  // Compiler error

f = (float)dec; // No error (May loss precision) ok

**dec = (decimal)dbl;** // No error  why it requires ?  

为什么上面的代码需要显式转换。浮点数/双精度数 > 十进制数的范围？

Answer 1

因为这不仅仅是关于精度；您也需要考虑范围- 坦率地说，1.79769313486232e308它真的非常大（如小数点左侧 > 300 位）。您的断言“小数的偶数范围 > 浮点数或双精度数”是不正确的。

var dbl = double.MaxValue;
var dec = (decimal) dbl; // BOOM!

doublea的范围大于a的范围decimal。

此外，您可能需要考虑double.NaN和。double.PositiveInfinitydouble.NegativeInfinity

Answer 2

我相信转换取决于数据类型的大小而不是数据类型的范围。

Integral Types

DataType    Size    . Net (CTS) Comments
byte    1   System.Byte 0 - 255 It is Unsigned
sbyte   1   System.SByte    -128 to 127 - Signed
short   2   System.Int16    
ushort  2   System.UInt16   
int     4   System.Int32     
uint    4   System.Unt32    
long    8   System.Int64    
ulong   8   System.UInt64   

Floating Types
decimal 16  System.Decimal  Has up to 28 digits after decimal
float   4   System.Single   Has up to 8 digits after decimal
double  8   System.Double   Has up to 15 digits after decimal