Question

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.

Answer 1

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.