C# LINQ Background Topics

IEnumerable<T> - Exiting a generator

There is a certain set of problems for which an unbounded generator is a great solution. But the majority of real-world situations deal with bounded problems. In fact, in order to use the GetFibonacci() method in practical examples, we found that it was necessary to impose some bounds on the sequence.

Exiting the generator method

One way to make a generator bounded is to simply allow the method to exit normally. Suppose we wanted GetFibonacci() to provide only the first n values of the sequence. We could write it like this:

private IEnumerable<int> GetFibonacciOfLength(int length)
{
    int previousVal1 = 0;
    int previousVal2 = 1;

    for (int i = 0; i < length; i++)
    {
        int nextVal = previousVal1 + previousVal2;
        previousVal1 = previousVal2;
        previousVal2 = nextVal;
        yield return nextVal;
    }
}

Now, we could use this new GetFibonacciOfLength() method like so:

// Will print:
// Value: 1
// Value: 2
// Value: 3
// Value: 5
// Value: 8
foreach (int val in GetFibonacciOfLength(5))
{
    Console.WriteLine($"Value: {val}");
}

Unlike with the previous examples, we don't need to break out of the foreach loop. It will complete normally at the end of the 5-value sequence.

The yield break statement

Suppose we want to have a generator that provides a bounded sequence, but it's non-trivial to calculate exactly how many values will be in the sequence. There is a way to do this as well.

Let's write a GetFibonacci() method that returns a sequence in which the maximum returned value is less-than-or-equal-to a provided max parameter. Here is one implementation of such a method:

private IEnumerable<int> GetFibonacciUpTo(int max)
{
    int previousVal1 = 0;
    int previousVal2 = 1;

    while (true)
    {
        int nextVal = previousVal1 + previousVal2;
        if (nextVal > max) yield break;
        previousVal1 = previousVal2;
        previousVal2 = nextVal;
        yield return nextVal;
    }
}

When execution hits the yield break statement, the iterator ends, as shown here using the Count() LINQ method:

// Prints the number 11, indicating that
// there are 11 values <= 200:
// 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

Console.WriteLine(GetFibonacciUpTo(200).Count());

Exercise

Exercise time again! Let's see if you can write a bounded generator. In this exercise, you will implement the same generator as in the unbounded generator exercise, but this time the sequence will be limited to the first n values, where n is a parameter to the method. As a reminder, here is the sequence you will be implementing:

Alternate between adding 2 to the previous number, then multiplying the previous number by 2. Here are the first several values in this sequence:

  0+2, 2x2, 4+2, 6x2, 12+2, 14x2, 28+2, 30x2, 60+2 ...
= 2,   4,   6,   12,  14,   28,   30,   60,   62 ...