生成均匀的随机浮点数，该浮点数可以返回所有可能的值

Question

An easy way to generate a random float64 in [0,1) is by generating a uniformly random int in [0,2⁵³) and dividing it by 2⁵³. This is essentially what rand.Float64() is doing. However, not all possible float64 values between 0 and 1 can be generated this way: if the value is lower than 2⁻⁴ for example, the 4 last bits of the significand are always going to be 0. Or, put more simply, the naive method always returns multiples of 2⁻⁵³, and not all floating point numbers between 0 and 1 are multiples of 2⁻⁵³.

How do you generate a uniformly random float64 such as every possible value has a chance of being returned? (Here, uniformly random means over the real interval [0,1): conceptually, I want to pick a uniformly random real number between 0 and 1 and return the closest float.)

For context, I need this because I'm implementing this paper and the assumption "all possible values between 0 and 1 are represented" is essential for the result to hold.

Answer 1

Porting this code (suggested in Severin's answer) is a possible option.

I think that it is equivalent to first generate the significand bits (by generating a random float in [1,2)), and then choose the exponent from a geometric distribution (it has a 0.5 chance of being -1, 0.25 of being -2, etc.).

// uniform returns a uniformly random float in [0,1).
func uniform() float64 {
  sig := rand.Uint64() % (1 << 52)
  return (1 + float64(i)/(1<<52)) / math.Pow(2, geometric())
}

// geometric returns a number picked from a geometric
// distribution of parameter 0.5.
func geometric() float64 {
  b := 1
  for rand.Uint64()%2 == 0 {
     b++
  }
  return b
}

We can probably make geometric() faster by using one of the LeadingZeros* functions from the bits package instead of doing one coin flip per bit.