Converting exponential number to decimal 1.11111117E+9 - trailing digits become zero

Question

I'm trying to convert and exponential number 1.11111117E+9 which is actually a 10 digit number '1111111111'. When I'm trying to convert this exponential number using decimal.TryParse method it is making last 3 digits as zero and giving the number as '111111000'. This is happening with any 10 digit number.

   decimal amount;
   decimal.TryParse("1.11111117E+9", NumberStyles.Any, null, out amount);

This is weird but I'm not able to figure out what's the issue here, can anybody tell me what's wrong in this?

Edit: Sorry for the misleading question. As Henrik mentioned in his answer is exactly what I'm facing.

float f = 1111111111;
string s = f.ToString();
decimal amount;
decimal.TryParse(s, NumberStyles.Any, null, out amount);

This will always return 1111111000? How do I address this issue to get the correct value? Change it to Double or Decimal from float datatype is the solution or anything else?

Answer 1

It's easy to round-trip a float value...

float f = 1111111111;
string s = f.ToString("r"); // r = roundtrip
float g = float.Parse(s);

Now f and g will be the same... but that doesn't mean that either value is exactly 1111111111... because that value can't be represented as a float. The nearest value exactly representable as a float is 1111111168... which is why you're getting the 7 at the end of the scientific representation.

Basically, you shouldn't be using float for this in the first place. From the docs for System.Single:

By default, a Single value contains only 7 decimal digits of precision, although a maximum of 9 digits is maintained internally.

So trying to store a 10 digit number and expecting it to be stored exactly is a fool's errand.

It's hard to say whether you should be using double or decimal - it depends on what value you're really trying to store. If it's a "natural" value like weight or height, I'd go for double. If it's a "human" value like a price, I'd use decimal.