C # retains 2 decimal places

1. Only ask to keep n bit not four 5 in

float F = 0.55555f;
int i = (int) (f * 100);
f = (float) (i*1.0)/100;
2. Keep n bits, rounded.

Decimal d= Decimal. Round (decimal. Parse ("0.55555"), 2);

3. Keep n-bit rounding

Math.Round (0.55555,2)

4, keep n-bit rounded
Double dbdata = 0.55555;
String str1 = Dbdata. ToString ("F2");//fn reserved n bits, rounded

5. Keep n-bit rounding


string result = String.Format ("{0:n2}", 0.55555);//2 bit

string result = String.Format ("{0:n3}", 0.55555);//3-bit

6. Keep n-bit rounding


Double s=0.55555;
Result=s.tostring ("#0.00");//point after 0 to keep several

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.

C # retains 2 decimal places