Function from Integers to the [0,1] float interval

Question

Given an integer I would like to produce a unique floating point number in the interval [0,1]. (this number will be used as id). The problem I found with all functions I have thought about, is that they encounter duplicates before running out of integer values. For example if f(a:int):float = 0.a then f(16000) = 0.16 and f(16001) = 0.16001. But since it is floating point, 0.16 and 0.16001 may be represented the same. In other words, I need a function that produce not only unique numbers, but also numbers that are represented uniquely (at least forthe C++ integer domain).

I know the answer is dependent on the size of integer and floating point in a specific environment, but if you can give an example for specific sizes it will still be helpful.

Answer 1

As someone else pointed out you can simply cast an int to a float of the same size to get a unique float (with some post-filtering for NaN and -0 and Inf). However, that will not meet your requirement of being in [0,1]. In fact, you can use that relation to show that there are not enough floats in [0,1] to represent the set of integers. If you use double then the mantissa is easily large enough for a 32-bit int and an expression like I / (double)INT_MAX should be sufficient (obviously allow for unsignedness if you need to).

Answer 2

If 23 bits are enough, you could build a float from int n like this:

float f;
*((__int32*)&f) = (0x7E<<23) | (n & 0x7FFFFF); // exponent=0x7E-0x7F=-1, fraction=n

It will map integers [0, 2^23-1] into floats [0.5, 1.0) according to the IEEE floating point standard.

If 23 bits are not enough, you have a solution at the link that you cited yourself. There are some other important notes, by the way, be sure to understand all the limitations.