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Rounding floats with `E` in C#

I've got 7,0975401565468943E+22
And Math.Round(x, 3) returns 7,0975401565468943E+22
Is it normal behavior and should I check if number contains E and if so just use something alike ToString("N2"); ?
code example:
float flo = float.Parse( " 7,0975401565468943E+22 " );
double flox = Math.Round(flo, 3);

The behavior you describe appears normal (though without a code example it is impossible to know for sure).
Your number has no significant digits in the fractional portion, as displayed. Note the "E+22", this means that you are dealing with a very large number. There are only 17 significant digits displayed, with another 5 digits not shown before you get to the decimal point. You can round to as many fractional digits as you want, you won't see any change in the number being displayed.

Parsing floats with Single.Parse()

Here comes a silly question. I'm playing with the parse function of System.Single and it behaves unexpected which might be because I don't really understand floating-point numbers. The MSDN page of System.Single.MaxValue states that the max value is 3.402823e38, in standard form that is
340282300000000000000000000000000000000
If I use this string as an argument for the Parse() method, it will succeed without error, if I change any of the zeros to an arbitrary digit it will still succeed without error (although it seems to ignore them looking at the result). In my understanding, that exceeds the limit, so What am I missing?

It may be easier to think about this by looking at some lower numbers. All (positive) integers up to 16777216 can be exactly represented in a float. After that point, only every other integer can be represented (up to the next time we hit a limit, at which point it's only every 4th integer that can be represented).
So what has to happen then is the 16777218 has to stand for 16777218∓1, 16777220 has to stand for 16777220∓1, etc. As you move up into even larger numbers, the range of integers that each value has to "represent" grows wider and wider - until the point where 340282300000000000000000000000000000000 represents all numbers in the range 340282300000000000000000000000000000000∓100000000000000000000000000000000, approximately (I've not actually worked out what the right ∓ value is here, but hopefully you get the point)
Number Significand Exponent
16777215 = 1 11111111111111111111111 2^0 = 111111111111111111111111
16777216 = 1 00000000000000000000000 2^1 = 1000000000000000000000000
16777218 = 1 00000000000000000000001 2^1 = 1000000000000000000000010
^
|
Implicit leading bit

That's actually not true - change the first 0 to 9 and you will see an exception. Actually change it to anything 6 and up and it blows up.
Any other number is just rounded down as float is not an 100% accurate representation of a decimal with 38+1 positions that's fine.

A floating point number is not like a decimal. It comprises a mantissa that carries the significant digits and an exponent that effectively says how far left or right of the decimal point to place the mantissa. A System.Single can only handle seven significant digits in the mantissa. If you replace any of your trailing zeroes with an arbitrary digit it is being lost when your decimal is converted into the mantissa and exponent form.

Good question. That is happening because the fact you can save a number with that range doesn't mean this type'll have enough precision to hold it. You can only store ~6-7 leading digits for floats and add an exponent to describe decimal point position.
0.012345 and 1234500 hold the same amount of informations - same mantissa, different exponents. The MSDN states only that value AFTRER EXPONENTIATION cannot be bigger, than MaxValue.

Rounding value to .5

There are the obvious quirks of Math.Round but is there a way to make Math.Round fulfill this type of manipulation for a rating system.
If greater than .0 and less than or equal to .5, display half a star
If greater than .5 and less than or equal to .0 display whole star
So obviously a half star would be .5 and a whole start would be the next whole value.
I don't know of a rounding method to go to half whole numbers.
Should I just write if statements to control my rounding?
**Edit/Solution
From the below answer I came up with.
double roundedRating = (Math.Ceiling(2 * currentRating)) / 2;

I'd recommend multiplying by two, performing Math.Ceiling, and dividing by two to get to the nearest half.

You're going to want to make sure that you end up performing your checks against integers, rather than floating point numbers.
Start by multiplying the number by 2. Continue doing this until it's an integer value (no value in the decimal part).
Now, continuously divide by 2 until you end up with a number that's less than or equal to the original number. If the result decimal part is greater than .0 and less than or equal to .5, display half a star. If it's greater than .5 and less than or equal to +.0, display a whole star.
Actually, go with matt's answer. ; )

Can this work?
Multiply the number by 10 e.g. 0.1x10, 0.2x10 to get n
Math.Ceil(n / 5) / 2
where n = 1, 2, 3 instead of - .1, .2, .3
examples:
1,2,3,4,5 = 1/2 = 0.5
6,7,8,9,10 = 2/2 = 1
11,12,13,14,15 = 3/2 = 1.5

If efficiency is no issue, the following approach could be used:
Number *= 10;
Number % 10 = remainder;
if(remainder <=5 && remainder != 0)
//Half star
else
//Whole star
However, that code is kinda ugly, but I think it gives you the general idea.

double roundedRating = (int)(currentRating * 2D + 0.999999999999999) / 2D

Unexpected Behavior of Math.Floor(double) and Math.Ceiling(double)

This question is about the threshold at which Math.Floor(double) and Math.Ceiling(double) decide to give you the previous or next integer value. I was disturbed to find that the threshold seems to have nothing to do with Double.Epsilon, which is the smallest value that can be represented with a double. For example:
double x = 3.0;
Console.WriteLine( Math.Floor( x - Double.Epsilon ) ); // expected 2, got 3
Console.WriteLine( Math.Ceiling( x + Double.Epsilon) ); // expected 4, got 3
Even multiplying Double.Epsilon by a fair bit didn't do the trick:
Console.WriteLine( Math.Floor( x - Double.Epsilon*1000 ) ); // expected 2, got 3
Console.WriteLine( Math.Ceiling( x + Double.Epsilon*1000) ); // expected 4, got 3
With some experimentation, I was able to determine that the threshold is somewhere around 2.2E-16, which is very small, but VASTLY bigger than Double.Epsilon.
The reason this question came up is that I was trying to calculate the number of digits in a number with the formula var digits = Math.Floor( Math.Log( n, 10 ) ) + 1. This formula doesn't work for n=1000 (which I stumbled on completely by accident) because Math.Log( 1000, 10 ) returns a number that's 4.44E-16 off its actual value. (I later found that the built-in Math.Log10(double) provides much more accurate results.)
Shouldn't the threshold should be tied to Double.Epsilon or, if not, shouldn't the threshold be documented (I couldn't find any mention of this in the official MSDN documentation)?

Shouldn't the threshold should be tied to Double.Epsilon
No.
The representable doubles are not uniformly distributed over the real numbers. Close to zero there are many representable values. But the further from zero you get, the further apart representable doubles are. For very large numbers even adding 1 to a double will not give you a new value.
Therefore the threshold you are looking for depends on how large your number is. It is not a constant.

The value of Double.Epsilon is 4.94065645841247e-324. Adding or subtracting this value to 3 results in 3, due to the way floating-point works.
A double has 53 bits of mantissa, so the smallest value you can add that will have any impact will be approximately 2^53 time smaller than your variable. So something around 1e-16 sounds about right (order of magnitude).
So to answer your question: there is no "threshold"; floor and ceil simply act on their argument in exactly the way you would expect.

This is going to be hand-waving rather than references to specifications, but I hope my "intuitive explanation" suits you well.
Epsilon represents the smallest magnitude that can be represented, that is different from zero. Considering the mantissa and exponent of a double, that's going to be extremely tiny -- think 10^-324. There's over three hundred zeros between the decimal point and the first non-zero digit.
However, a Double represents roughly 14-15 digits of precision. That still leaves 310 digits of zeros between Epsilon and and integers.
Doubles are fixed to a certain bit length. If you really want arbitrary precision calculations, you should use an arbitrary-precision library instead. And be prepared for it to be significantly slower -- representing all 325 digits that would be necessary to store a number such as 2+epsilon will require roughly 75 times more storage per number. That storage isn't free and calculating with it certainly cannot go at full CPU speed.

C# float infinite loop

The following code in C# (.Net 3.5 SP1) is an infinite loop on my machine:
for (float i = 0; i < float.MaxValue; i++) ;
It reached the number 16777216.0 and 16777216.0 + 1 is evaluates to 16777216.0. Yet at this point: i + 1 != i.
This is some craziness.
I realize there is some inaccuracy in how floating point numbers are stored. And I've read that whole numbers greater 2^24 than cannot be properly stored as a float.
Still the code above, should be valid in C# even if the number cannot be properly represented.
Why does it not work?
You can get the same to happen for double but it takes a very long time. 9007199254740992.0 is the limit for double.

Right, so the issue is that in order to add one to the float, it would have to become
16777217.0
It just so happens that this is at a boundary for the radix and cannot be represented exactly as a float. (The next highest value available is 16777218.0)
So, it rounds to the nearest representable float
16777216.0
Let me put it this way:
Since you have a floating amount of precision, you have to increment up by a higher-and-higher number.
EDIT:
Ok, this is a little bit difficult to explain, but try this:
float f = float.MaxValue;
f -= 1.0f;
Debug.Assert(f == float.MaxValue);
This will run just fine, because at that value, in order to represent a difference of 1.0f, you would need over 128 bits of precision. A float has only 32 bits.
EDIT2
By my calculations, at least 128 binary digits unsigned would be necessary.
log(3.40282347E+38) * log(10) / log(2) = 128
As a solution to your problem, you could loop through two 128 bit numbers. However, this will take at least a decade to complete.

Imagine for example that a floating point number is represented by up to 2 significant decimal digits, plus an exponent: in that case, you could count from 0 to 99 exactly. The next would be 100, but because you can only have 2 significant digits that would be stored as "1.0 times 10 to the power of 2". Adding one to that would be ... what?
At best, it would be 101 as an intermediate result, which would actually be stored (via a rounding error which discards the insignificant 3rd digit) as "1.0 times 10 to the power of 2" again.

To understand what's going wrong you're going to have to read the IEEE standard on floating point
Let's examine the structure of a floating point number for a second:
A floating point number is broken into two parts (ok 3, but ignore the sign bit for a second).
You have a exponent and a mantissa. Like so:
smmmmmmmmeeeeeee
Note: that is not acurate to the number of bits, but it gives you a general idea of what's happening.
To figure out what number you have we do the following calculation:
mmmmmm * 2^(eeeeee) * (-1)^s
So what is float.MaxValue going to be? Well you're going to have the largest possible mantissa and the largest possible exponent. Let's pretend this looks something like:
01111111111111111
in actuality we define NAN and +-INF and a couple other conventions, but ignore them for a second because they're not relevant to your question.
So, what happens when you have 9.9999*2^99 + 1? Well, you do not have enough significant figures to add 1. As a result it gets rounded down to the same number. In the case of single floating point precision the point at which +1 starts to get rounded down happens to be 16777216.0

It has nothing to do with overflow, or being near the max value. The float value for 16777216.0 has a binary representation of 16777216. You then increment it by 1, so it should be 16777217.0, except that the binary representation of 16777217.0 is 16777216!!! So it doesn't actually get incremented or at least the increment doesn't do what you expect.
Here is a class written by Jon Skeet that illustrates this:
DoubleConverter.cs
Try this code with it:
double d1 = 16777217.0;
Console.WriteLine(DoubleConverter.ToExactString(d1));
float f1 = 16777216.0f;
Console.WriteLine(DoubleConverter.ToExactString(f1));
float f2 = 16777217.0f;
Console.WriteLine(DoubleConverter.ToExactString(f2));
Notice how the internal representation of 16777216.0 is the same 16777217.0!!

The iteration when i approaches float.MaxValue has i just below this value. The next iteration adds to i, but it can't hold a number bigger than float.MaxValue. Thus it holds a value much smaller, and begins the loop again.