Double.MIN_VALUE和Double.MAX_VALUE之间是否是随机的两倍? [英] Random double between Double.MIN_VALUE and Double.MAX_VALUE?

问题描述

任何人都知道如何实现这一目标.我已经尝试了通常的公式，但是我只得到正数< = 10:

Anyone got an idea of how to achieve this. I've tried the usual formula but I'm only getting positive numbers <= 10:

Double.MIN_VALUE + Math.random() * ((Double.MAX_VALUE - Double.MIN_VALUE) + 1)

推荐答案

您可以这样做

private static final Random rand = new Random();

public static double getRandomDouble() {
    while(true) {
        double d = Double.longBitsToDouble(rand.nextLong());
        if (d < Double.POSITIVE_INFINITY && d > Double.NEGATIVE_INFINITY)
            return d;
    }
}

这将以相等的概率返回任何有限的double.

This will return any finite double with equal probability.

随着(Double.MAX_VALUE - (-Double.MAX_VALUE))溢出到无穷大，您不能只是上面的公式.即所有正负双精度值的范围都太大，无法存储为双精度值.

You can't just the the formula above as the (Double.MAX_VALUE - (-Double.MAX_VALUE)) overflows to infinity. i.e. the range for all positive and negative double values is too large to store in a double.

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