Why is the Convert.ToDouble (string) round?

Question

For example, Convert.ToDouble ("0.1234567890123456789") = 0.123456789012346

What is the maximum number of significant digits? Could not find it in the documents.

Answer 1

From MSDN :

Accuracy: 15-16 digits.

Answer 2

Of course, there is maximum accuracy. This is the maximum that you can express with the bits used to store the double. For this line, you can try Decimal instead.

Answer 3

, . IEEE 754-1985 .

,

sign * 1.mantissa * 2^(exponent - bias)

, , .

Answer 4

http://en.wikipedia.org/wiki/Double_precision

The double has only 64 bits of storage, and some of them are used for sign and exponent, so for the fractional part there is a limited number of digits (usually 15). A float has only 32 bits, therefore 8-9 digits of accuracy. Also, some digits cannot be accurately represented using a floating point number, so you can see some rounding errors in your conversions.