using System;
using System.Linq;
using System.Collections.Generic;

public class FunctionalStyleProgramming
{
	private static void PrintList(IEnumerable<int> list)
	{
		foreach(int elem in list){
			Console.Write(elem);
			Console.Write(' ');
		}
		Console.WriteLine();
	}
	
	/// Problem: You need to get a list that contains even numbers
	///          ranged from 0 up to 100
	/// 
	public static void Main()
	{
		// First, using the traditional for-loop
		var even_numbers_for = new List<int>();
		even_numbers_for.Capacity = 100;
		for(int i = 0; i < 100; ++i){
			if(i % 2 == 0)
				even_numbers_for.Add(i);
		}
		
		// Second, using the slightly sophisticated foreach-loop
		var even_numbers_foreach = new List<int>();
		even_numbers_foreach.Capacity = 100;
		foreach(int j in Enumerable.Range(0, 100)){
			if(j % 2 == 0)
				even_numbers_foreach.Add(j);
		}
		
		// Third, using LINQ to Objects
		var even_numbers_lto =
			Enumerable.Range(0, 100)
					  .Where(n => n % 2 == 0);
		
		// And lastly using LINQ to SQL(which I strongly recommend to use, and is familiar to those who are used to SQL)
		var even_numbers_lts =
			from n in Enumerable.Range(0, 100)
			where n % 2 == 0
			select n;
			//Darn Ideone! It doesn't hightlight LINQ to SQL syntaxes! I'm very disappointed at that ;-(
			
		PrintList(even_numbers_for);
		PrintList(even_numbers_foreach);
		PrintList(even_numbers_lto);
		PrintList(even_numbers_lts);
		// And then think about the pros and cons of each solution.
		// Which one do you think is the best solution?
		//
		// Extra...
		Note();
	}

	private static void Note()
	{
		var first_even_number =
			(from n in Enumerable.Range(1, 100)
			 where n % 2 == 0
			 select n
			).FirstOrDefault();

		Console.WriteLine("The first even number above 0 is {0}.", first_even_number);

		var recommended =
			from n in Enumerable.Range(1, 100)
			where n % 2 == 0
			select n;

		var recommended_first_even_number = recommended.FirstOrDefault();
		Console.WriteLine("The first even number above 0 is {0}.", recommended_first_even_number);
	}
}