Question

有没有办法迭代这个循环并且：

var resAndNumber = new Dictionary<int, int>();
for (int i = 0; i < numberRepeats; i++)
{

    for (int n = 0; n < numberDice; n++)
    {

        rolledResult1 += diceRandomGenerate.Next(1, 7);
    }

    //if the sum was not thrown, entry in the dictionary
    if (!resAndNumber.ContainsKey(rolledResult1))
    {
        resAndNumber[rolledResult1] = 0;
    }

    // bisherige  Anzahl für diese Summe hochzählen
    resAndNumber[rolledResult1]++;
}

让控制台的输出可以用一个foreach完成，我正在尝试了解。在我陷入死胡同之前，我仍然对基础知识很古怪 - 你能给我任何建议吗？

为了满足问题/目标，允许用户决定要模拟多少骰子。谢天谢地，c＃starter。

The output may look like this

Answer 1

Dictionary<T,S>实现了IEnumerable<KeyValuePair<T, S>>，因此可以枚举它：

foreach (var kv in dic) {
  Console.WriteLine($"({kv.Key}, {kv;.Value})");
}

但是，跟踪每个结果的滚动数量的整个想法可以通过数组更容易地跟踪（其中条目将默认为零）：

var results = new int[6*diceCount + 1];
// Will ignore indicies 0 to diceCount-1 as easier than using
// non-zero based arrays.

for (var roll = 1; roll <= rollCount; ++roll) {
  var rollResult = Enumerable.Range(0, diceCount)
                             .Select(x => diceRandomGenerate.Next(1, 7))
                             .Sum();
  results[rollResult]++;
}

for (var roll = diceCount; roll <= diceCount*6; ++roll) {
  Console.WriteLine($"{roll:d2} " + new String('*', results[roll]));
}

Answer 2

这样的东西？：

foreach (var keyValuePair in resAndNumber)
{
    Console.WriteLine($"Die eye {keyValuePair.Key} for the 1. die was {keyValuePair.Value} ties thrown");
}

请注意，您不会存储未滚动的结果，结果将按照首次滚动的顺序排列。因此，如果你滚动1,6,3,6,1,3,3，结果将是（1,2），（6,2）和（3,3）。

您可能希望首先填写字典：

var resAndNumber = new Dictionary<int, int> {{1, 0}, {2, 0}, {3, 0}, {4, 0}, {5, 0}, {6, 0}};

然后你可以检查结果是否已经在字典中。

示例：

var resAndNumber = new Dictionary<int, int> {{1, 0}, {2, 0}, {3, 0}, {4, 0}, {5, 0}, {6, 0}};

resAndNumber[1]++;
resAndNumber[6]++;
resAndNumber[3]++;
resAndNumber[6]++;
resAndNumber[4]++;

foreach (var keyValuePair in resAndNumber)
{
    Console.WriteLine($"Die eye {keyValuePair.Key} for the 1. die was {keyValuePair.Value} ties thrown");
}

通过字典迭代foreach

2 个答案: