I am trying to understand how to round to the nearest tenths position with C#. For instance, I have a value that is of type double. This double is currently set to 10.75. However, I need to round and then truncate everything past the tenths position. In this case, I am seeking a value of 10.8. How do I round to the tenths position in C#?
Thank you!
Math.Round(yourNumber, 1)
The second parameter is number of decimal places to round to. In your case you want 1 decimal place as an end result.
You simply need to use the overload of Math.Round that takes the decimals parameter.
Math.Round(10.75, 1) // returns 10.8
Just for comparison:
Math.Round(10.75) // returns 11
Math.Round(10.75, 0) // returns 11
Math.Round(10.75, 2) // returns 10.75
Since you Used Math.Round() in your title, I'm going to assume you've already tried the basic Math.Round(10.75,1) approach and it returns something you don't expect. With that in mind, I suggest looking at some of the different overloads for the function, specifically one that accepts a MidPointRounding enum:
http://msdn.microsoft.com/en-us/library/f5898377.aspx
Do you really need to round it, or can you just format it for printing but allow the variable itself to hold its precision? Something like:
decimal value = 10.75;
value.ToString ("#.#");
If you just want to "cut" everything after the first decimal, this shoudl work :
return Math.Round(value * 10)/10
Related
Can somebody please tell me what is math.ceiling function doing here because I think Math.Pow will do the job without it? You don't need the rest of the code to understand because the code would do the same without Math.Ceiling. I just need information whether Math.Ceiling makes sure that Math.Pow has really calculated its number in the right way or does it mean something else. Thanks in advance
a[i] = a[i-1] * 10 + 45 * (int)(Math.Ceiling(Math.Pow(10, i-1)));
The call is completely pointless. Since i is an index into the array, it must be a positive integer - otherwise, this will either crash (if i is 0 or less) due to not being a valid index into the array or not even compile in the first place (if i is not an integer at all). That being said, Math.pow is already guaranteed to return an integer in this case, so it will never make any difference in the output.
Math.ceiling() is the guarantee that your number will be an integer, like, if the pow returns a float number, Math.ceiling will round up the number.
Ex.: Pow returns 2.7, Math.ceiling returns 3.
Assuming that i is already an integer, there is no real need for the Math.ceiling() function, because the pow return will always be an integer.
Hope that is the answer you are looking for.
I have the double value like 12.256852651 and I want to display it as 12.257 as a float number without converting it in to a string type.
How can I do it in C# ?
I'd first convert to Decimal and then use Math.Round on the result. This conversion is not strictly necessary, but I always feel a bit uneasy if I round to decimal places while using binary floating points.
Math.Round((Decimal)f, 3, MidpointRounding.AwayFromZero)
You should also look into the choice of MidpointRounding, since by default this uses Banker's round, which is not what you are used to from school.
If you want to display it, it will be a string and that's what you need to use.
If you want to round in order to use it later in calculations, use Math.Round((decimal)myDouble, 3).
If you don't intend to use it in calculation but need to display it, use double.ToString("F3").
how to make the Rounded number ?
Example : 3341.48 to 3342.00
It seems you always want to round up here. In that case use
Math.Ceiling(3341.48)
This will return 3342.
If you want to round towards the nearest whole number, use
Math.Round(3341.48)
This will return 3341. Note that Bankers roundingis the default setting here, that might cause some unexpected result for rounding X.50.
If you want 3341.48 to round up to 3342, it sounds like you might want Math.Ceiling:
decimal m = 3341.48m;
decimal roundedUp = Math.Ceiling(m);
This will always round up - so 3341.0000001 would still round to 3342, for example. If that's not what you're after, please specify the circumstances in which you want it to round up, and those in which you want it to round down instead.
Note that this will round up to 3342, not 3342.00 - it doesn't preserve the original precision, because you've asked for an integer value by using Math.Ceiling.
It's relatively unusual to then want to force the precision to 2, but you could divide by 100 and then multiply by 100 again, if necessary. Alternatively, if you only need this for output you should look into formatting the result appropriately rather than changing the value.
Use Math.Round(number) if you want to round number to the nearest integer.
Use Math.Round(number,digits) if you want to round number to a specified number of fractional digits.
If you want to round to lower/higer value use Math.Floor(number) / Math.Ceiling(number) instead.
To round monetary amounts to 5 cents:
amount = 20 * int(amount / 20)
I am using Convert.ChangeType() to convert from Object (which I get from DataBase) to a generic type T. The code looks like this:
T element = (T)Convert.ChangeType(obj, typeof(T));
return element;
and this works great most of the time, however I have discovered that if I try to cast something as simple as return of the following sql query
select 3.2
the above code (T being double) wont return 3.2, but 3.2000000000000002. I can't realise why this is happening, or how to fix it. Please help!
What you're seeing is an artifact of the way floating-point numbers are represented in memory. There's quite a bit of information available on exactly why this is, but this paper is a good one. This phenomenon is why you can end up with seemingly anomalous behavior. A double or single should never be displayed to the user unformatted, and you should avoid equality comparisons like the plague.
If you need numbers that are accurate to a greater level of precision (ie, representing currency values), then use decimal.
This probably is because of floating point arithmetic. You probably should use decimal instead of double.
It is not a problem of Convert. Internally double type represent as infinite fraction of 2 of real number, that is why you got such result. Depending of your purpose use:
Either Decimal
Or use precise formating {0:F2}
Use Math.Flor/Math.Ceil
I'm wondering if it's possible for .Net's String.Format() to split an integer apart into two sub strings. For example I have a number 3234 and I want to format it as 32X34. My integer will always have 4 or 6 digits. Is this possible using String.Format()? If so what format string would work?
P.S.
I know there is other ways to do this, but i'm specifically interested to know if String.Format() can handle this.
You can specify your own format when calling String.Format
String.Format("{0:00x00}", 2398) // = "23x93"
James, I'm not sure you've completely specified the problem. If your goal is to put the 'x' in the center of the string, Samuel's answer won't work for 6 digit numbers. String.Format("{0:00x00}", 239851) returns "2398x51" instead of "239x851"
Instead, try:
String.Format(val<10000 ? "{0:00x00}" : "{0:000x000}", val)
In either case, the method is called Composite Formatting.
(I'm assuming the numbers will be between 1000 and 999999 inclusive. Even then, numbers between 1000 and 1009 inclusive will report the number after the 'x' with an unnecessary leading '0'. So maybe this approach is valid for values between 1010 and 999999 inclusive.)
No, it can't.
In fact, it seems that your integers aren't integers. Perhaps they should be stored in a class, with its own ToString() method that will format them this way.