产生一个随机数，如果数字与之前相匹配 [英] Generate a random number if the number matched with the previous

问题描述

我想创建一个在它前面文字的随机数，而我不希望显示数量的两倍系统。所以，这是我做的方式：

I want to create a random number with a text in front of it, and I don't want the system displaying a number twice. So, this is that the way that I did it:

Random _rand = new Random();

private void RandomNumberGenerator()
        {
            int random = _rand.Next(10000);
            string text = "TP0" + random;

            if (random.Equals(random))
            {
                _rand.Next();
            }

            else
            {
                random = _rand.Next(10000);
            }

            MessageBox.Show(text);
        }

我没有得到任何显示数量的两倍（但我也不太清楚，因为我刚关闭程序，直到它显示的号码5次（这一切都不相同数量）。

I am not getting any displayed number twice (but I am not too sure, because I just close the program until it displayed the number 5 times (all of it are not the same number).

是否有可能从上面的代码来显示一个数与任何机会，两次？

Is it possible from the above code to displaying a number twice with any chance?

感谢您。

推荐答案

编辑以。摆脱幻数，并确保理智

EDITED to get rid of magic numbers and ensure sanity.

  Random _rand = new Random();
  HashSet<int> _taken = new HashSet<int>();
  object _syncRoot = new object();

  private int RandomNumberGenerator() {
    lock (_syncRoot) {
      const int MAX_NUMBER = 10000;
      if (_taken.Count == MAX_NUMBER) {
        throw new Exception("All possible numbers are already generated.");
      }

      int random = _rand.Next(MAX_NUMBER);
      while (_taken.Contains(random)) {
        random = (random + 1) % MAX_NUMBER;
      }
      _taken.Add(random);
      return random;
    }
  }

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