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1. 与运营商一道制作飞行的假件[]
原标题:Create Dictionary item on the fly with operator[]

通常,当你创建<代码>Dictionary<Tkey, TValue>时,你必须首先去,通过在字典上添加字句。

我有<代码>Dictionary<string, mycontainer>,其中是其他物体的集装箱。 我需要能够很快在座标上添加一些内容,这样我就认为,如果飞机已经存在,我可以超负荷使用(<代码>operator[],在飞行上打上“/code>,然后允许我直接增加:

mydictionnary [“SomeName”].Add(myobject);,未明文要求每当字典上存在一个有上述名称的集装箱时,就建造我的集装箱。

我想知道,这是否是一个好的想法,或者我是否应当明确制造新的我的物体?

最佳回答

您应使自己的班子达到<代码>。 Dictionary<TKey, List<TItem>>。

索引员希望:

public List<TItem> this[TKey key] {
    get {
        List<TItem> retVal;
        if (!dict.TryGetValue(key, out retVal))
            dict.Add(key, (retVal = new List<TItem>(itemComparer)));
        return retVal;
    }
}
问题回答

Might as well do myCustomClass.Add ( key, subkey, Value ); so that the Code can be accessible and intellisense will Guide it s use.

我要谈的是明确的法典,即索引员的间接超负荷实际上不是一个共同的情况,在几个月里,其他人将读一下你的法典,这很可能是 b的。





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