C
C#•7mo ago
eid

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
Pobiega•7mo 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
LasseVK•7mo 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
cap5lut•7mo 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
MODiX•7mo 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
cap5lut•7mo 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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