IEnumerable<T> provides two GetEnumerator methods - what is the difference between them?

落花浮王杯 提交于 2019-12-22 04:29:09

问题


When I emplement IEnumerable<T> interface I see two GetEnumerator methods: one returning IEnumerator and other IEnumerator<T>. When would I use one or another?


回答1:


If you are implementing the IEnumerable<T> generic interface, you will pretty much always have to use the generic GetEnumerator method - unless you cast your object explicitly to (non-generic) IEnumerable.

The reason is backwards compatability with .NET 1.0/1.1 which didn't support generics.




回答2:


You usually implement both. One is the newer, generic version that returns a typesafe enumerator (IEnumerator<T>). The other one is for compatibility with Legacy code (returns IEnumerator). A typical implementation is:

public IEnumerator<T> GetEnumerator() {
    foreach(T item in items) {
        yield return item;
    }
}

IEnumerator IEnumerable.GetEnumerator() {
    return GetEnumerator();
}



回答3:


The reason there are two methods is because IEnumerable<T> inherits the IEnumerable interface so you are seeing the generic method from IEnumerable<T> and the non-generic method from IEnumerable.

Here is how you want to implement the interface in your type:

class Foo : IEnumerable<Foo>
{
    public IEnumerator<Foo> GetEnumerator()   
    {
        // do your thing here
    }

    // do a EIMI here and simply call the generic method
    IEnumerator IEnumerable.GetEnumerator()
    {
        this.GetEnumerator();
    }
}



回答4:


Usually GetEnumerator() calls GetEnumerator<T>(), so there should not be much difference in behavior. As for why there are two methods, this is done for backwards compatibility and for use in situation where T is not of great interest (or is just unknown).




回答5:


The one is generic, the other one not. I believe the compiler prefers to use the generic overload.



来源:https://stackoverflow.com/questions/560678/ienumerablet-provides-two-getenumerator-methods-what-is-the-difference-betwe

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