Why is the "f" required when declaring floats?

C#Floating Point

C# Problem Overview


Example:

float timeRemaining = 0.58f;

Why is the f is required at the end of this number?

C# Solutions


Solution 1 - C#

Your declaration of a float contains two parts:

  1. It declares that the variable timeRemaining is of type float.
  2. It assigns the value 0.58 to this variable.

The problem occurs in part 2.

The right-hand side is evaluated on its own. According to the C# specification, a number containing a decimal point that doesn't have a suffix is interpreted as a double.

So we now have a double value that we want to assign to a variable of type float. In order to do this, there must be an implicit conversion from double to float. There is no such conversion, because you may (and in this case do) lose information in the conversion.

The reason is that the value used by the compiler isn't really 0.58, but the floating-point value closest to 0.58, which is 0.57999999999999978655962351581366... for double and exactly 0.579999946057796478271484375 for float.

Strictly speaking, the f is not required. You can avoid having to use the f suffix by casting the value to a float:

float timeRemaining = (float)0.58;

Solution 2 - C#

Because there are several numeric types that the compiler can use to represent the value 0.58: float, double and decimal. Unless you are OK with the compiler picking one for you, you have to disambiguate.

The documentation for double states that if you do not specify the type yourself the compiler always picks double as the type of any real numeric literal:

> By default, a real numeric literal on the right side of the assignment > operator is treated as double. However, if you want an integer number > to be treated as double, use the suffix d or D.

Appending the suffix f creates a float; the suffix d creates a double; the suffix m creates a decimal. All of these also work in uppercase.

However, this is still not enough to explain why this does not compile:

float timeRemaining = 0.58;

The missing half of the answer is that the conversion from the double 0.58 to the float timeRemaining potentially loses information, so the compiler refuses to apply it implicitly. If you add an explicit cast the conversion is performed; if you add the f suffix then no conversion will be needed. In both cases the code would then compile.

Solution 3 - C#

The problem is that .NET, in order to allow some types of implicit operations to be carried out involving float and double, needed to either explicitly specify what should happen in all scenarios involving mixed operands or else allow implicit conversions between the types to be performed in one direction only; Microsoft chose to follow the lead of Java in allowing the direction which occasionally favors precision, but frequently sacrifices correctness and generally creates hassle.

In almost all cases, taking the double value which is closest to a particular numeric quantity and assigning it to a float will yield the float value which is closest to that same quantity. There are a few corner cases, such as the value 9,007,199,791,611,905; the best float representation would be 9,007,200,328,482,816 (which is off by 536,870,911), but casting the best double representation (i.e. 9,007,199,791,611,904) to float yields 9,007,199,254,740,992 (which is off by 536,870,913). In general, though, converting the best double representation of some quantity to float will either yield the best possible float representation, or one of two representations that are essentially equally good.

Note that this desirable behavior applies even at the extremes; for example, the best float representation for the quantity 10^308 matches the float representation achieved by converting the best double representation of that quantity. Likewise, the best float representation of 10^309 matches the float representation achieved by converting the best double representation of that quantity.

Unfortunately, conversions in the direction that doesn't require an explicit cast are seldom anywhere near as accurate. Converting the best float representation of a value to double will seldom yield anything particularly close to the best double representation of that value, and in some cases the result may be off by hundreds of orders of magnitude (e.g. converting the best float representation of 10^40 to double will yield a value that compares greater than the best double representation of 10^300.

Alas, the conversion rules are what they are, so one has to live with using silly typecasts and suffixes when converting values in the "safe" direction, and be careful of implicit typecasts in the dangerous direction which will frequently yield bogus results.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionThomasView Question on Stackoverflow
Solution 1 - C#Jeffrey SaxView Answer on Stackoverflow
Solution 2 - C#JonView Answer on Stackoverflow
Solution 3 - C#supercatView Answer on Stackoverflow