我写了
但是dieOne和dieTwo总是相等的，您如何制作两个独立的随机值? Mikanu ，现在可以使用了.
在2009年7月13日星期一10:43 PM修改解决方案
您应创建一个Random实例作为类的静态字段，并在需要时使用它.它永远不应是方法的局部变量.
正确的方法是创建 Random 类的实例，然后调用以下方法之一:-Next(Int32 max) // will return a random number smaller than max -NextByte() // will return a random byte (between 0 and 255) -NextDouble() // will return a random number between 0.0 and 1.0<br /> 这是一个代码示例，它将生成介于0和10之间的两个随机数a和b <br />Random rnd = new Random();<br />int a = rnd.Next(10);<br />int b = rng.Next(10);<br />
I wrote
<br /> Random RollOne = new Random();<br /> Random RollTwo = new Random();<br /> dieOne = RollOne.Next(1, 7);<br /> dieTwo = RollTwo.Next(1 , 7);<br /> sum = dieOne + dieTwo;<br />but dieOne and dieTwo are always equal, how do you make two independent random values?Edit:Thanks Mikanu, it works now.
modified on Monday, July 13, 2009 10:43 PM 解决方案 You should create one instance of Random as a static field of the class and use it whenever you need it.It should never be a local variable of a method.The correct way to do that is to create an instance of the Random class and then call one of the following methods: - Next(Int32 max) // will return a random number smaller than max - NextByte() // will return a random byte (between 0 and 255) - NextDouble() // will return a random number between 0.0 and 1.0<br />Here''s a code sample which will generate two random numbers, a and b, between 0 and 10<br />Random rnd = new Random();<br />int a = rnd.Next(10);<br />int b = rng.Next(10);<br />