How to print 1 to 100 without any looping using C#

Question

I am trying to print numbers from 1 to 100 without using loops, using C#. Any clues?

Answer 1

Console.WriteLine('1');
Console.WriteLine('2');
...
Console.WriteLine('100');

...Or would you have accepted a recursive solution?

EDIT: or you could do this and use a variable:

int x = 1;
Console.WriteLine(x);
x+=1;
Console.WriteLine('2');
x+=1;
...
x+=1
Console.WriteLine('100');

Answer 2

No loops, no recursion, just a hashtable-like array of functions to choose how to branch:

using System;
using System.Collections.Generic;

namespace Juliet
{
    class PrintStateMachine
    {
        int state;
        int max;
        Action<Action>[] actions;

        public PrintStateMachine(int max)
        {
            this.state = 0;
            this.max = max;
            this.actions = new Action<Action>[] { IncrPrint, Stop };
        }

        void IncrPrint(Action next)
        {
            Console.WriteLine(++state);
            next();
        }

        void Stop(Action next) { }

        public void Start()
        {
            Action<Action> action = actions[Math.Sign(state - max) + 1];
            action(Start);
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            PrintStateMachine printer = new PrintStateMachine(100);
            printer.Start();
            Console.ReadLine();
        }
    }
}

Answer 3

class Program
{
    static Timer s = new Timer();
    static int i = 0;
    static void Main(string[] args)
    {
        s.Elapsed += Restart;
        s.Start();
        Console.ReadLine();
    }
    static void Restart(object sender, ElapsedEventArgs e)
    {
        s.Dispose();
        if (i < 100)
        {
            Console.WriteLine(++i);
            s = new Timer(1);
            s.Elapsed += Restart;
            s.Start();
        }
    }
}

You must notice that I'm NOT using recursion.

Answer 4

The cool and funny way:

static void F(int[] array, int n)
{
    Console.WriteLine(array[n] = n);
    F(array, n + 1);
}
static void Main(string[] args)
{
    try { F(new int[101], 1); }
    catch (Exception e) { }
}

Answer 5

By the time I answer this, someone will already have it, so here it is anyway, with credit to Caleb:

void Main()
{
    print(0, 100);
}

public void print(int x, int limit)
{
    Console.WriteLine(++x);
    if(x != limit)
        print(x, limit);
}

Answer 6

    [Test]
    public void PrintNumbersNoLoopOrRecursionTest()
    {
        var numberContext = new NumberContext(100);

        numberContext.OnNumberChange += OnNumberChange(numberContext);
        numberContext.CurrentNumber = 1;
    }

    OnNumberChangeHandler OnNumberChange(NumberContext numberContext)
    {
        return (o, args) =>
        {
            if (args.Counter > numberContext.LastNumber)
                return;

            Console.WriteLine(numberContext.CurrentNumber);

            args.Counter += 1;
            numberContext.CurrentNumber = args.Counter;
        };
    }


public delegate void OnNumberChangeHandler(object source, OnNumberChangeEventArgs e);
public class NumberContext
{
    public NumberContext(int lastNumber)
    {
        LastNumber = lastNumber;
    }

    public event OnNumberChangeHandler OnNumberChange;
    private int currentNumber;
    public int CurrentNumber
    {
        get { return currentNumber; }
        set {
            currentNumber = value;
            OnNumberChange(this, new OnNumberChangeEventArgs(value));
        }
    }

    public int LastNumber { get; set; }

    public class OnNumberChangeEventArgs : EventArgs
    {
        public OnNumberChangeEventArgs(int counter)
        {
            Counter = counter;
        }

        public int Counter { get; set; }
    }