How C # retains the number of decimal places

Http://www.cnblogs.com/furenjun/archive/2010/07/13/1776484.html

Double dvalue = 95.12345;

int ivalue = 10000;
string strvalue = "95.12345";
string result = "";

result = Convert.todouble (Dvalue). ToString ("0.00");//two digits after the decimal point, with a result of 95.12
result = Convert.todouble (ivalue). ToString ("0.00");//10000.00
result = Convert.todouble (strvalue). ToString ("0.00");//95.12

result = Convert.todouble (Dvalue). ToString ("P");//Get a percentage of 2 digits after the decimal point, plus the% number automatically;//9512.35%
result = Convert.todouble (strvalue).  ToString ("F4");//reserved 4 digits after decimal point; 95.1235
One thing to note is that convert.todouble must be this double-precision, otherwise it will be an error.

How C # retains the number of decimal places