C#•4mo ago

enumerable.range vs yield return

i can't understand the different bw the iterator that has made by yield return and other iterators(normal iterator, enumerable.range iterator)

5 Replies

Pobiega•4mo ago

Whats the confusion? you could in theory implement Enumerable.Range using yield return as such:

public static class MyEnumerable
{
    public static IEnumerable<int> Range(int start, int count)
    {
        for (int i = start; i < start + count; i++)
        {
            yield return i;
        }
    }
}

public static class MyEnumerable
{
    public static IEnumerable<int> Range(int start, int count)
    {
        for (int i = start; i < start + count; i++)
        {
            yield return i;
        }
    }
}

its really not more complicated than that

LasseVK•4mo ago

What exactly is the question here? Might it be that you want to understand how a method implemented using yield return actually works?

cap5lut•4mo ago

first of all they are called enumerators in c#/.net (linq has iterators, buts thats basically an implementation detail for them) for the difference: for yield return stuff the compiler will generate its own type that implements the IEnumerator<T> and more. thats quite a bit of code thats actually generated if u look at

MODiX•4mo ago

cap5lut

sharplab.io (click here)

public class C {
    public IEnumerable<int> M() {
        for (int i = 0; i < 10; i++) {
            yield return i;
        }
    }
}

public class C {
    public IEnumerable<int> M() {
        for (int i = 0; i < 10; i++) {
            yield return i;
        }
    }
}

React with ❌ to remove this embed.

cap5lut•4mo ago

so when using Enumerable.Range() instead, the compiler would not create a new type for it, which is quite beneficial long term because not every library will have its own type to do the same thing

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enumerable.range vs yield return