Let's explore how extension methods in C# can solve the expression problem. For an introduction to the expression problem, refer to my first post in this series. Extension methods allow operations to be defined over a type without modifying the type's original definition.

First, we define IExpr as a marker interface and implement it in the Const and Add types. The implementations of Const and Add here are identical to those in the previous post.

public interface IExpr { }

public class Const : IExpr
{
    public double Value { get; }

    public Const(double value) =>
        Value = value;
}

public class Add : IExpr
{
    public IExpr Left { get; }
    public IExpr Right { get; }

    public Add(IExpr left, IExpr right) =>
        (Left, Right) = (left, right);
}

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Programmers are always defining types and operations to use these types. It's the essence of developing features in working software. The expression problem^[1] asks how easy it is to define types and operations in a given programming language or paradigm, and is stated as follows:

"The goal is to define a datatype by cases, where one can add new cases to the datatype and new functions over the datatype, without recompiling existing code, and while retaining static type safety (e.g., no casts)."

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Let me just say that I am not particularly a fan of either microbenchmarks or premature optimization. I also feel that LINQ extensions are a fine addition to the .NET standard library. The LINQ query syntax is also an integral part of the C# and F# languages.

That being said, there's an interesting and revealing tale to be told about the performance of LINQ to Objects.

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Functional languages have the notion of lazy sequences, which are an abstraction of infinite sequences that are stored using a small, finite amount of memory. It would be wasteful to realize an entire infinite sequence before even using it. The basic idea is to only call the function that generates the sequence when needed, and cache the results. With lazy sequences, you don't blow the stack and the elements in the sequence are not recalculated everytime.

Let's look at how the two most popular and functional JVM languages handle lazy sequences.

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