C# Random.Next 总是返回相同的值
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Random.Next returns always the same values
提问by Thomas
This is really weird, and I cannot see why this is happening. In the foreach cycle, I am iterating through a class A collection, and for each class, I call the Count()method, where r1and r2numbers are generated from range [-1,1]. The problem is that Random.Nextreturns the same "random" numbers for each instance. When the results for the first instance are 0 and -1, the same ones will be returned from following instances. Please, could you tell me why this is happening? Also, I cannot get different results in each class A instance. This is the code:
这真的很奇怪，我不明白为什么会这样。在foreach循环，我迭代通过类的集合，并且对于每个类，我调用Count()方法，其中，r1和r2数据是从范围[-1,1]生成。问题是 Random.Next为每个实例返回相同的“随机”数字。当第一个实例的结果为 0 和 -1 时，后续实例将返回相同的结果。拜托，你能告诉我为什么会这样吗？此外，我无法在每个 A 类实例中获得不同的结果。这是代码：
class a
{
Random rnd = new Random();
private void Count()
{
int r1 = rnd.Next(-1, 1);
int r2 = rnd.Next(-1, 1);
}
}
class b
{
List<a> listofA=new list<a>();
foreach (a ACLASS in listofA)
{
ACLASS.Count();
}
}
采纳答案by Guffa
The problem is that you are creating instances of the Randomclass too close in time.
问题是您创建的Random类的实例时间太近了。
When you create a Randomobject, it's seeded with a value from the system clock. If you create Randominstances too close in time, they will all be seeded with the same random sequence.
当您创建一个Random对象时，它会从系统时钟中获得一个值。如果您创建的Random实例时间太近，它们将全部以相同的随机序列进行播种。
Create a single Randomobject and pass its reference to the constructor when you create instances of the "a" class, instead of creating one Randomobject for each "a" instance.
在创建Random“a”类的实例时，创建一个对象并将其引用传递给构造函数，而不是Random为每个“a”实例创建一个对象。
回答by Jherico
You include a random instance for each A instance. It sounds like they're all getting the same default seed value. You probably want to make a static random for all A instances and use it repeatedly, or alternatively provide a seed value to the Random() instance in the A constructor.
您为每个 A 实例包含一个随机实例。听起来他们都获得了相同的默认种子值。您可能希望为所有 A 实例创建一个静态随机数并重复使用它，或者为 A 构造函数中的 Random() 实例提供一个种子值。
回答by ChrisF
You're creating a new instance of Randomvery close together (your loop is very tight) so each instance is effectively using the same seed value.
您正在创建一个Random非常接近的新实例（您的循环非常紧密），因此每个实例都有效地使用了相同的种子值。
A better approach would be to create one instance and pass that to your Countmethod.
更好的方法是创建一个实例并将其传递给您的Count方法。
You problably know this next bit, but I'll include it here for completeness:
您可能知道下一点，但为了完整起见，我将其包括在此处：
The MSDNhas the details on this, but basically your problem is the Random.Nextmethod you're using generates:
在MSDN对这个细节，但基本上你的问题是Random.Next你使用生成方法：
A 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not maxValue. If minValue equals maxValue, minValue is returned.
大于或等于 minValue 且小于 maxValue 的 32 位有符号整数；即返回值的范围包括minValue，但不包括maxValue。如果 minValue 等于 maxValue，则返回 minValue。
because of this your calls will return -1 or 0.
因此，您的调用将返回 -1 或 0。
回答by tvanfosson
Use a single, static Random number generator for all instances of the class.
对类的所有实例使用单个静态随机数生成器。
class a
{
private static Random rnd;
static a() {
rnd = new Random();
}
private void Count()
{
int r1 = rnd.Next(-1, 2);
int r2 = rnd.Next(-1, 2);
}
}
Note the change to give you numbers in the range -1,1 rather than -1,0
请注意为您提供 -1,1 而不是 -1,0 范围内的数字的更改
