foreach是最快的，如果只迭代___个值，它也会更快

在.NET Framework 4.7中，可以使用分解

var fruits = new Dictionary<string, int>();
...
foreach (var (fruit, number) in fruits)
{
    Console.WriteLine(fruit + ": " + number);
}

要使此代码在较低的C#版本上运行，请添加System.ValueTuple NuGet包并在某处编写

public static class MyExtensions
{
    public static void Deconstruct<T1, T2>(this KeyValuePair<T1, T2> tuple,
        out T1 key, out T2 value)
    {
        key = tuple.Key;
        value = tuple.Value;
    }
}

foreach是最快的，如果只迭代___个值，它也会更快

我写了一个扩展来遍历字典。

public static class DictionaryExtension
{
    public static void ForEach<T1, T2>(this Dictionary<T1, T2> dictionary, Action<T1, T2> action) {
        foreach(KeyValuePair<T1, T2> keyValue in dictionary) {
            action(keyValue.Key, keyValue.Value);
        }
    }
}

然后你可以打电话

myDictionary.ForEach((x,y) => Console.WriteLine(x + " - " + y));

C#7.0引入了解构器，如果您正在使用.NET Core 2.0+应用程序，那么结构KeyValuePair＜＞已经为您提供了一个解构器（）。因此，您可以做到：

var dic = new Dictionary<int, string>() { { 1, "One" }, { 2, "Two" }, { 3, "Three" } };
foreach (var (key, value) in dic) {
    Console.WriteLine($"Item [{key}] = {value}");
}
//Or
foreach (var (_, value) in dic) {
    Console.WriteLine($"Item [NO_ID] = {value}");
}
//Or
foreach ((int key, string value) in dic) {
    Console.WriteLine($"Item [{key}] = {value}");
}

foreach(KeyValuePair<string, string> entry in myDictionary)
{
    // do something with entry.Value or entry.Key
}