下應該做的和它重用釋放了鍵:
internal class AutoKeyDictionary<TKey, TValue> : IEnumerable<KeyValuePair<TKey, TValue>>, IEnumerable
{
private readonly Dictionary<TKey, TValue> inner;
private readonly Func<TKey, TKey> incrementor;
private readonly Stack<TKey> freeKeys;
private readonly TKey keySeed;
private TKey currentKey;
public AutoKeyDictionary(TKey keySeed, Func<TKey, TKey> incrementor)
{
if (keySeed == null)
throw new ArgumentNullException("keySeed");
if (incrementor == null)
throw new ArgumentNullException("incrementor");
inner = new Dictionary<TKey, TValue>();
freeKeys = new Stack<TKey>();
currentKey = keySeed;
}
public TKey Add(TValue value) //returns the used key
{
TKey usedKey;
if (freeKeys.Count > 0)
{
usedKey = freeKeys.Pop();
inner.Add(usedKey, value);
}
else
{
usedKey = currentKey;
inner.Add(usedKey, value);
currentKey = incrementor(currentKey);
}
return usedKey;
}
public void Clear()
{
inner.Clear();
freeKeys.Clear();
currentKey = keySeed;
}
public bool Remove(TKey key)
{
if (inner.Remove(key))
{
if (inner.Count > 0)
{
freeKeys.Push(key);
}
else
{
freeKeys.Clear();
currentKey = keySeed;
}
return true;
}
return false;
}
public bool TryGetValue(TKey key, out TValue value) { return inner.TryGetValue(key, out value); }
public TValue this[TKey key] { get {return inner[key];} set{inner[key] = value;} }
public bool ContainsKey(TKey key) { return inner.ContainsKey(key); }
public bool ContainsValue(TValue value) { return inner.ContainsValue (value); }
public int Count { get{ return inner.Count; } }
public Dictionary<TKey,TValue>.KeyCollection Keys { get { return inner.Keys; } }
public Dictionary<TKey, TValue>.ValueCollection Values { get { return inner.Values; } }
public IEnumerator<KeyValuePair<TKey, TValue>> GetEnumerator() { return inner.GetEnumerator(); }
IEnumerator IEnumerable.GetEnumerator() { return ((IEnumerable)inner).GetEnumerator(); }
}
聲明:我沒有測試此代碼,它可能有一點重要的幾個pesty錯誤,一般的做法是合理的。
只需用你描述的方法編寫一個包裝字典的類 - 我假設你知道你將如何從你的對象中生成你的密鑰? –
鍵不會從對象中生成,鍵將用於將對象關聯在一起。我試着循環每個鍵,並檢查這個鍵是否等於0,如果鍵0不存在,我會用該鍵創建新的對象,如果存在,我會重複1,2,3,4 ...但我知道這會造成很大的性能問題。 – None
新密鑰是如何產生的?如果它只是自動遞增,那麼你可以包裝一個'List'。 –
Lee