Implementing Iterators in C # (Part 2)

Implementing Iterators in C # (Part 1) .

Now that you have in your luggage a general idea of ​​what is behind the iterators, you can already answer some questions about their use. Here is a scenario based on real events:

I have a large and complex iterator, and I want to refactor it. For clarity, we agree that the enumerator counts from 1 to 100 twice. (Of course, in real life, the iterator will not be so simple.)

IEnumerable CountTo100Twice()
{
    int i;
    for (i = 1; i <= 100; i++) {
        yield return i;
    }
    for (i = 1; i <= 100; i++) {
        yield return i;
    }
}

As we know from Pragramming 101, we can put the common code into a subroutine and call it. But when I do this, I get a compiler error:

IEnumerable CountTo100Twice()
{
    CountTo100();
    CountTo100();
}
void CountTo100()
{
    int i;
    for (i = 1; i <= 100; i++) {
        yield return i;
    }
}

What am I doing wrong? How do I transfer “counting to 100” to a subroutine and call it twice from the CountTo100Twice function?

As we just saw, iterators are not routines at all. The technique presented above would work perfectly if we built iterators from, say, fibers instead of finite state machines. But since automata are used, yield return expressions should be located at the “top level”. So how will you iterate through routines?

You make the subroutine itself an iterator, and then “pull” the results from the main function:

IEnumerable CountTo100Twice()
{
    foreach (int i in CountTo100()) yield return i;
    foreach (int i in CountTo100()) yield return i;
}
IEnumerable CountTo100()
{
    for (i = 1; i <= 100; i++) {
        yield return i;
    }
}

Exercise: Consider the following snippet:

foreach (int i in CountTo100Twice()) {
    // ...
}

Explain what happens on the 150th (fifty) call to MoveNext () in the loop above. Consider the implications of this for recursive algorithms (such as tree traversal).