java java中介于0.0和1.0之间的随机浮点数
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Random float between 0.0 and 1.0 both inclusive in java
提问by rickbear
Possible Duplicate:
Random floating point double in Inclusive Range
可能的重复：
包含范围内的随机浮点双精度
I'm working on a project where I need the random float between 0.0 and 1.0 both inclusive. After considering a while I found out that it's propably not that important for my project anyway, so I just use the standard:
我正在做一个项目，我需要在 0.0 和 1.0 之间的随机浮点数。考虑了一段时间后，我发现无论如何它对我的项目来说可能并不那么重要，所以我只使用标准：
new Random().nextFloat();
which generates a float from 0.0 (inlcusive) to 1.0 (exclusive).
它生成从 0.0（包含）到 1.0（不包含）的浮点数。
But I can't stop thinking how it's actually accomplished to make with 1.0 inclusive?
但我不能停止思考它是如何用 1.0 inclusive 完成的？
One method I could think of is (pseudocode):
我能想到的一种方法是（伪代码）：
new Random().nextFloat() * (1.0 + "lowest possible value above 0");
Another method I could think of is:
我能想到的另一种方法是：
float myRandomFloat = 0;
float rand1 = new Random().nextFloat(); // [0;1)
float rand2 = 1.0f - new Random().nextFloat(); // (0;1]
boolean randPicker = new Random().nextBoolean();
if (randPicker)
myRandomFloat = rand1;
else
myRandomFloat = rand2;
But I think this ought to be more simple or not? So my real question is: Is there a real or better way of doing this?
但我认为这应该更简单吗？所以我真正的问题是：有没有真正的或更好的方法来做到这一点？
回答by Euclides Mulémbwè
if you multiply Random().nextFloat() by 10 it will give you a number from 0 to 9 which are the base numbers in the numeral system, so number 10 doesn't really exist but it's a combination of two base numbers, so you will never be able to get it as a random number, but if you really need it I would go with your second option.
如果你将 Random().nextFloat() 乘以 10，它会给你一个从 0 到 9 的数字，这是数字系统中的基数，所以数字 10 并不真正存在，但它是两个基数的组合，所以您将永远无法将其作为随机数获取，但如果您真的需要它，我会选择您的第二个选项。
