C#中System.Random类的NextDouble()方法用于返回大于或等于0.0且小于1.0的随机浮点数。
用法:
public virtual double NextDouble();
返回值:此方法返回一个大于或等于0.0且小于1.0的双精度浮点数。
以下示例程序旨在说明NextDouble()方法的使用:
示例1:
// C# program to illustrate the
// Random.NextDouble() Method
using System;
class GFG {
// Driver code
public static void Main()
{
// Instantiate random number generator
Random rand = new Random();
// Print 10 random floating point numbers
Console.WriteLine("Printing 10 random floating point numbers");
for (int i = 0; i < 10; i++)
Console.WriteLine("{0} -> {1}", i, rand.NextDouble());
}
}
输出:
Printing 10 random floating point numbers 0 -> 0.0227202852362396 1 -> 0.624568469647583 2 -> 0.0145442797870116 3 -> 0.646489209330869 4 -> 0.967497945748036 5 -> 0.839329582098559 6 -> 0.873648912121378 7 -> 0.16200648022909 8 -> 0.66018275761054 9 -> 0.0837694853934317
示例2:
// C# program to illustrate the
// Random.NextDouble() Method
using System;
class GFG {
// Driver code
public static void Main()
{
// Instantiate random number generator
Random rand = new Random();
// Instantiate an array of double
double[] a = new double[10];
// Store random floating point
// numbers in the array
for (int i = 0; i < 10; i++)
a[i] = rand.NextDouble();
// Print 10 random floating point numbers
Console.WriteLine("Printing 10 random "+
"floating point numbers");
for (int i = 0; i < 10; i++)
Console.WriteLine("{0} -> {1}", i, a[i]);
}
}
输出:
Printing 10 random floating point numbers 0 -> 0.853536825558886 1 -> 0.741455778359182 2 -> 0.496043408986201 3 -> 0.0975164361752181 4 -> 0.120282317567748 5 -> 0.57163705703413 6 -> 0.749181974562435 7 -> 0.684014179596684 8 -> 0.691246760865323 9 -> 0.888019556127498
参考:
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注:本文由纯净天空筛选整理自rupesh_rao大神的英文原创作品 C# | Random.NextDouble() Method。非经特殊声明,原始代码版权归原作者所有,本译文未经允许或授权,请勿转载或复制。