Question

我有一个字典，它使用ulong作为键，结构作为值，即

public Dictionary<UInt64, OptimalOutputs> What_to_do>

最佳输出结构

public struct OptimalOutputs
    {
        public short gotoschool;
        public short goForDining;
        public short GoToAcademy;

    }

如何在字典中迭代打印，每个键和值一起？我尝试了keyvalue对但是徒劳无功

Answer 1

foreach(KeyValuePair<UInt64, OptimalOutputs> pair in dict)
{
    Console.WriteLine("Key: {0} Values: {1},{2},{3}", 
        pair.Key,  
        pair.Value.gotoschool, pair.Value.goForDining, pair.Value.GoToAcademy);
}

您还可以在结构中覆盖ToString，这有助于f.e.调试：

public struct OptimalOutputs
{
    public short GotoSchool;
    public short GoForDining;
    public short GoToAcademy;

    public override string ToString()
    {
        return string.Format("{0}, {1}, {2}", GoForDining, GotoSchool, GoToAcademy);
    }
}

现在您可以使用这个较短的版本（隐式调用ToString）：

foreach (KeyValuePair<UInt64, OptimalOutputs> pair in dict)
{
    Console.WriteLine("Key: {0} Values: {1}", pair.Key, pair.Value);
}

Answer 2

您只需使用foreach次迭代：

foreach (KeyValuePair<UInt64, OptimalOutputs> pair in dictionary)
{
     Console.WriteLine(string.Format("Key: {0} ", pair.Key);
     Console.WriteLine(string.Format("Values: {0}, {1}, {2}", 
                                 pair.Value.goForDining,  
                                 pair.Value.gotoschool,                      
                                 pair.Value.GoToAcademy);
}

<强> 其他：

顺便说一句，您可以覆盖ToString()模型的OptimalOutputs。

public class OptimalOutputs
{
    ...
    public override string ToString() 
      {
         return string.Format("Values: {0}, {1}, {2}", goForDining, gotoschool, GoToAcademy);
      }
    ...
}

然后：

foreach (KeyValuePair<UInt64, OptimalOutputs> pair in dictionary)
{
     Console.WriteLine(string.Format("Key: {0} Values: {1}", pair.Key, pair.Value);
}

在c＃中打印字典键值对

2 个答案: