How to round decimal value after .50 in C#?

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I want to round the decimal value like this
if 5.50 => 5
if 5.51 => 6 ..
I want if the decimal value after point getter than .50 then the value increment by 1.
I used Math.Round() function, but it's round after .49 decimal value..

Posted 21-Feb-13 0:31am

JayantaChatterjee

Updated 25-Jan-17 2:35am

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Philippe Mori 25-Jan-17 12:40pm

Using non standard rounding seems suspicious...

4 solutions

Solution 1

basic and simple way , you can convert decimal (5.50) to int (5) and you can check for 5.50-5 greater or lower than .50 if bigger then apply your algorithim..but i dont know any other function does that automaticly.

Posted 21-Feb-13 0:37am

boogac

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JayantaChatterjee 21-Feb-13 6:44am

how to get after point(.) values?

boogac 21-Feb-13 6:47am

decimal yourDecimal=5.50;
int temp = (int)yourDecimal;//or Convert.toInt32(..) 5.50 will be 5
decimal pointAfterDecimal= yourDecimal-(decimal)temp;

Solution 2

try,
String.Format("{0: 0}", _decimal_value);

hopes this work for you.

Posted 21-Feb-13 0:41am

Hrushikesh_phapale

Updated 21-Feb-13 1:05am

v2

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Mitchell J. · Accepted Answer · 2013-02-21T00:42:00

Solution 3

Use a custom rounding system:

C#

public int Round(double value)
{
    double decimalpoints = Math.Abs(value - Math.Floor(value));
    if (decimalpoints > 0.5)
        return (int)Math.Round(value);
    else
        return (int)Math.Floor(value);
}

Hope this helps. :-)

Posted 21-Feb-13 0:42am

Mitchell J.

Comments

JayantaChatterjee 21-Feb-13 6:50am

Thanks a lotttttttttttt Sir..

CHill60 · Accepted Answer · 2013-02-21T00:48:00

Solution 4

C#

public int myRound(double val)
{
    return (int)Math.Floor(val + .49);
}

I don't like "magic numbers" so you might want to define a const double and swap it for the 0.49

Posted 21-Feb-13 0:48am

CHill60

Updated 21-Feb-13 0:50am

v2

Comments

JayantaChatterjee 21-Feb-13 6:55am

I think This is more efficient then previous answers...
Thanks Sir....
Thanks a Lottttttttttt...:-)

Philippe Mori 25-Jan-17 12:39pm

But it will not do the same thing for .495!

CHill60 27-Jan-17 10:50am

I think your comment means that it won't work for values such as 5.501 for example. True, point well made. The solution is to use 0.499 as the magic number, or as many 9's as you suspect you will have digits after the decimal point. Not nice.
Although the OP stated they found my solution more efficient, Solution 3 is the better answer and works regardless of the mantissa.

kurt minds 29-May-18 10:53am

use Format function. example Convert.toDouble(Format(5.4,"N0")) and it will return 6

CHill60 29-May-18 14:40pm

Which isn't what the OP wanted and includes an unnecessary conversion to a string. I also don't like Convert functions