我有一个HashTable,我跟踪颜色(这是键)和颜色的数量,这是关键。
我试图弄清楚当HashTable已经包含颜色时如何增加键。这是一段代码:
Hashtable htColors = new Hashtable();
if (htColors.Contains(color))
{
// Want to increase the "value" of the key here.
}
else
{
htColors.Add(color, 1); //Found color for first time
}
答案 0 :(得分:5)
我发布这个是迂腐的。我不喜欢与Dictionary的接口,因为这种非常常见的访问需要花费 - 如果你最常见的情况是触摸已经存在的元素,你必须哈希并查找你的值3次。不相信我?我在这里写了DK的解决方案:
static void AddInc(Dictionary<string, int> dict, string s)
{
if (dict.ContainsKey(s))
{
dict[s]++;
}
else
{
dict.Add(s, 1);
}
}
当放入IL时 - 你得到这个:
L_0000: nop
L_0001: ldarg.0
L_0002: ldarg.1
L_0003: callvirt instance bool [mscorlib]System.Collections.Generic.Dictionary`2<string, int32>::ContainsKey(!0)
L_0008: ldc.i4.0
L_0009: ceq
L_000b: stloc.0
L_000c: ldloc.0
L_000d: brtrue.s L_0028
L_000f: nop
L_0010: ldarg.0
L_0011: dup
L_0012: stloc.1
L_0013: ldarg.1
L_0014: dup
L_0015: stloc.2
L_0016: ldloc.1
L_0017: ldloc.2
L_0018: callvirt instance !1 [mscorlib]System.Collections.Generic.Dictionary`2<string, int32>::get_Item(!0)
L_001d: ldc.i4.1
L_001e: add
L_001f: callvirt instance void [mscorlib]System.Collections.Generic.Dictionary`2<string, int32>::set_Item(!0, !1)
L_0024: nop
L_0025: nop
L_0026: br.s L_0033
L_0028: nop
L_0029: ldarg.0
L_002a: ldarg.1
L_002b: ldc.i4.1
L_002c: callvirt instance void [mscorlib]System.Collections.Generic.Dictionary`2<string, int32>::Add(!0, !1)
L_0031: nop
L_0032: nop
L_0033: ret
调用ContainsKey,get_item和set_item,所有这些都散列并查找。
我写了一些不太漂亮的东西,它使用了一个包含int的类,并且该类允许你对它产生副作用(由于struct复制语义,你不能真正使用结构而不会产生同样的惩罚)。
class IntegerHolder {
public IntegerHolder(int x) { i = x; }
public int i;
}
static void AddInc2(Dictionary<string, IntegerHolder> dict, string s)
{
IntegerHolder holder = dict[s];
if (holder != null)
{
holder.i++;
}
else
{
dict.Add(s, new IntegerHolder(1));
}
}
这为您提供以下IL:
L_0000: nop
L_0001: ldarg.0
L_0002: ldarg.1
L_0003: callvirt instance !1 [mscorlib]System.Collections.Generic.Dictionary`2<string, class AddableDictionary.IntegerHolder>::get_Item(!0)
L_0008: stloc.0
L_0009: ldloc.0
L_000a: ldnull
L_000b: ceq
L_000d: stloc.1
L_000e: ldloc.1
L_000f: brtrue.s L_0023
L_0011: nop
L_0012: ldloc.0
L_0013: dup
L_0014: ldfld int32 AddableDictionary.IntegerHolder::i
L_0019: ldc.i4.1
L_001a: add
L_001b: stfld int32 AddableDictionary.IntegerHolder::i
L_0020: nop
L_0021: br.s L_0033
L_0023: nop
L_0024: ldarg.0
L_0025: ldarg.1
L_0026: ldc.i4.1
L_0027: newobj instance void AddableDictionary.IntegerHolder::.ctor(int32)
L_002c: callvirt instance void [mscorlib]System.Collections.Generic.Dictionary`2<string, class AddableDictionary.IntegerHolder>::Add(!0, !1)
L_0031: nop
L_0032: nop
L_0033: ret
调用get_item一次 - 在对象存在的情况下没有额外的散列。我有点肮脏,并将该字段公开,以避免方法调用属性访问。
如果是我,我会将这个整体功能包装到自己的类中,并将IntegerHolder类隐藏在公共视图中 - 这是一个限制版本:
public class CountableItem<T>
{
private class IntegerHolder
{
public int i;
public IntegerHolder() { i = 1; }
}
Dictionary<T, IntegerHolder> dict = new Dictionary<T, IntegerHolder>();
public void Add(T key)
{
IntegerHolder val = dict[key];
if (val != null)
val.i++;
else
dict.Add(key, new IntegerHolder());
}
public void Clear()
{
dict.Clear();
}
public int Count(T key)
{
IntegerHolder val = dict[key];
if (val != null)
return val.i;
return 0;
}
// TODO - write the IEnumerable accessor.
}
答案 1 :(得分:4)
尝试以下
if (htColors.Contains(color))
{
int old = (int)htColors[color];
htColor[color] = old + 1;
}
编辑对评论的回复
恕我直言,字典方法要好得多,因为它是1)类型安全和2)消除了此解决方案中涉及的拳击。
让行成为以下内容不会影响密钥,只会影响值
htColor[color] = (int)htColor[color] + 1;