Question

如何在一个范围内生成随机数？

Answer 1

你可以尝试

Random r = new Random();
int rInt = r.Next(0, 100); //for ints
int range = 100;
double rDouble = r.NextDouble()* range; //for doubles

看看

Random Class，Random.Next Method (Int32, Int32)和Random.NextDouble Method

Answer 2

尝试以下代码。

Random rnd = new Random();
int month = rnd.Next(1, 13); // creates a number between 1 and 12
int dice = rnd.Next(1, 7); // creates a number between 1 and 6
int card = rnd.Next(52); // creates a number between 0 and 51

Answer 3

类似的东西：

var rnd = new Random(DateTime.Now.Millisecond);
int ticks = rnd.Next(0, 3000);

Answer 4

使用：

Random r = new Random();
 int x= r.Next(10);//Max range

Answer 5

对于未来的读者，如果您想要一个范围内的随机数，请使用以下代码：

public double GetRandomNumberInRange(double minNumber, double maxNumber)
{
    return new Random().NextDouble() * (maxNumber - minNumber) + minNumber;
}

C# Random double between min and max

Code sample

Answer 6

除了生成整数和双精度的Random Class之外，请考虑：

Stack Overflow问题 Generation of (pseudo) random constrained values of (U)Int64 and Decimal
C# RandomProvider Class

Answer 7

这里是Darrelk answer的更新版本。使用C# extension methods实现。每次调用此方法时，它不会不分配内存（新的Random（））。

public static class RandomExtensionMethods
{
    public static double NextDoubleRange(this System.Random random, double minNumber, double maxNumber)
    {
        return random.NextDouble() * (maxNumber - minNumber) + minNumber;
    }
}

用法（确保导入包含RandomExtensionMethods类的名称空间）：

var random = new System.Random();
double rx = random.NextDoubleRange(0.0, 1.0);
double ry = random.NextDoubleRange(0.0f, 1.0f);
double vx = random.NextDoubleRange(-0.005f, 0.005f);
double vy = random.NextDoubleRange(-0.005f, 0.005f);

使用C＃在一个范围内生成一个随机数

7 个答案: