问题描述
我在寻找一种有效的方式来产生就开开区间（0,1）随机浮点数。我现在有上的封闭闭区间生成随机整数的RNG [0，（2 ^ 32）-1]。我已经通过简单地通过我的结果，从整数的RNG乘以创建在区间[0,1）半开浮点RNG 1 /（（2 ^ 32）-1），而不是除以（2 ^ 32） -1，因为它是低效率的。
I'm looking for an efficient way to generate random floating-point numbers on the open-open interval (0,1). I currently have an RNG that generates random integers on the closed-closed interval of [0, (2^32)-1]. I've already created a half-open floating point RNG on the interval [0,1) by simply multiplying my result from the integer RNG by 1/((2^32)-1) rather than dividing by (2^32)-1 since it's inefficient.
目前我打算约在区间（0,1）生成号码的方法是用类似下面的条件语句：
The way I'm currently going about generating numbers on the interval (0,1) is with a conditional statement like the one below:
float open_open_flt = (closed_open_flt==0) ? closed_open_flt : FLT_MIN;
不幸的是，这是相当低效的，因为它是控制code，我觉得它引入了一些偏差。
Unfortunately, this is rather inefficient since it is control code and I feel like it introduces some bias.
任何人都可以提出一个替代方案？
Can anybody suggest an alternative?
推荐答案
您已经在那里了。
两个浮点数当前的发电机产生之间的最小距离为1 /（2 ^ 32）。
The smallest distance between two floats your current generator produces is 1/(2^32).
所以，你的发电机efectively生产的 [0.1-1 /（2 ^ 32）]
So, your generator is efectively producing [0,1-1/(2^32)].
1 /（2 ^ 32）大于FLT_MIN更大。
1/(2^32) is greater than FLT_MIN.
因此，如果您添加FLT_MIN到您的发电机，
Thus if you add FLT_MIN to your generator,
float open_open_flt = FLT_MIN + closed_open_flt;
你会得到的 [FLT_MIN，1-（1 /（2 ^ 32））+ FLT_MIN] ，它可以作为一个（0,1）发电机。
you'll get [FLT_MIN,1-(1/(2^32))+FLT_MIN], which works as a (0,1) generator.
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