多个随机数是相同的 [英] Multiple random numbers are the same

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本文介绍了多个随机数是相同的的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述


可能重复:结果
随机数生成器只生成一个随机数






一个初学者的问题。我有一个非常简单的程序,绘制一条线,我想随机的位置,但每次我创建一个新的实例随机它返回相同的值。哪里有问题?谢谢



 私人无效抽奖()
{
随机random1 =新的随机();
INT randomNumber1 = random1.Next(0,300);
随机random2 =新的随机();
INT randomNumber2 = random2.Next(0,300);
随机random3 =新的随机();
INT randomNumber3 = random3.Next(0,300);
随机random4 =新的随机();
INT randomNumber4 = random4.Next(0,300);
System.Drawing.Graphics G = this.CreateGraphics();
朋绿=新朋(Color.Green,5);
g.DrawLine(绿色,新点(randomNumber1,randomNumber2),
新点(randomNumber3,randomNumber4));
}

私人无效btndraw1_Click(对象发件人,EventArgs五)
{
抽奖();
}


解决方案

发生这种情况的原因是,每次你做一个新的随机时间它正在使用的时钟初始化。因此,在一个紧密的循环(或多个电话一前一后),你得到的价值,很多时候一样,因为所有的随机变量都使用相同的种子初始化。



要解决这个问题:创建只有一个随机变量,最好之外的功能,并且只使用一个实例

 随机random1 =新的随机( ); 
私人无效抽奖()
{
INT randomNumber1 = random1.Next(0,300);
INT randomNumber2 = random1.Next(0,300);
INT randomNumber3 = random1.Next(0,300);
INT randomNumber4 = random1.Next(0,300);
System.Drawing.Graphics G = this.CreateGraphics();
朋绿=新朋(Color.Green,5);
g.DrawLine(绿色,新点(randomNumber1,randomNumber2),新点(randomNumber3,randomNumber4));
}


Possible Duplicate:
Random number generator only generating one random number

A beginner question. I have a very simple program that draws a line and I want to randomize the locations, but each time I create a new instance of Random it returns the same value. Where is the problem? Thank you.

private void Draw()
{
    Random random1 = new Random();
    int randomNumber1 = random1.Next(0, 300);
    Random random2 = new Random();
    int randomNumber2 = random2.Next(0, 300);
    Random random3 = new Random();
    int randomNumber3 = random3.Next(0, 300);
    Random random4 = new Random();
    int randomNumber4 = random4.Next(0, 300);
    System.Drawing.Graphics g = this.CreateGraphics();
    Pen green = new Pen(Color.Green, 5);
    g.DrawLine(green, new Point(randomNumber1, randomNumber2), 
                      new Point(randomNumber3, randomNumber4));
}

private void btndraw1_Click(object sender, EventArgs e)
{
    Draw();
}

解决方案

The reason this happens is that every time you do a new Random it is initialized using the clock. So in a tight loop (or many calls one after the other) you get the same value lots of times since all those random variables are initialized with the same seed.

To solve this: Create only one Random variable, preferably outside your function and use only that one instance.

Random random1 = new Random();
private void Draw()
{
    int randomNumber1 = random1.Next(0, 300);
    int randomNumber2 = random1.Next(0, 300);
    int randomNumber3 = random1.Next(0, 300);
    int randomNumber4 = random1.Next(0, 300);
    System.Drawing.Graphics g = this.CreateGraphics();
    Pen green = new Pen(Color.Green, 5);
    g.DrawLine(green, new Point(randomNumber1, randomNumber2), new Point(randomNumber3, randomNumber4));
}

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