问题描述
如何将从crypto.randomBytes返回的字节数组转换为浮点数？我写一个小的替换Math.random（）
How do I convert the array of bytes returned from crypto.randomBytes into a to a floating point number? I'm writing a small replacement for Math.random()
推荐答案
假设你有一系列字节随机选择均匀分布over [0,256）。取这些字节中的七个，例如，，... a 。计算（（（（（（a 6/32）/ 32 + a 5）/ 256 + a 4）/ 256 + a ）/ 256 + a ）/ 256 + a ）/ 256 + a 。
Suppose you have a series of bytes randomly selected with uniform distribution over [0, 256). Take seven of those bytes, say a, a,… a. Calculate (((((((a % 32)/32 + a)/256 + a)/256 + a)/256 + a)/256 + a)/256 + a)/256.
说明：％32表示模32的余数。这需要五位a 。然后除以32将这些位移位到小数点的右边，添加八个新位，除以256移位右8位，依此类推，直到我们有53位。
Explanation: a % 32 denotes the residue of a modulo 32. This takes five bits of a. Then division by 32 "shifts" these bits to the right of the radix point, eight new bits are added, a division by 256 "shifts" right eight bits, and so on until we have 53 bits.
这会在[0，1]中提供2个种可能的结果。可以提供更多，因为浮点分辨率越精细，当值越接近零，但是有均匀性和其他问题的问题。
This provides 2 possible results in [0, 1). It is possible to provide more since the floating-point resolution is finer as values get closer to zero, but then there are problems with uniformity and other issues.
这篇关于javascript中的crypto.randomBytes（）的浮点数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！