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Quote:

Stargazer, you with your head in the heavens,
You'll never get by walkin' that high off the ground.
Moon dreamer, I've been around and I've seen it;
The higher you get--the harder they let you down.
You pay your dues, it seems forever.
And if you're clever you may be in for a while,
Then you're out of style.

glennPattonWork wrote:

Guess what my companies favoured Goggle search is this PM

1. If the year is set to '00' in PHP, it apparently doesn't mean '0000', instead it is short hand for '2000'. So '00-00-00' means The year '2000', the month '00', and the day '00'.
2. Okay, but the month can't be zero! '00' means you have to subtract one from the year, which makes the year '1999', and the month '12'. That gets us to '1999-12-00'
3. Drat, day is also zero! Time to subtract a day from the month, '1999-11-30'

class Integer
{
    public int Value { get; set; }

    public Integer() =>
        Value = 0;

    public Integer(int number) =>
        Value = number;

    public static Integer operator +(Integer d1, Integer d2) =>
        new Integer(d1.Value + d2.Value);

    public static string operator +(string s, Integer d) =>
        s + d.Value;

    public static string operator +(Integer d, string s) =>
        d.Value + s;

    public static Integer operator -(Integer d1, Integer d2) =>
        new Integer(d1.Value - d2.Value);

    public static Integer operator -(Integer d, string s) =>
        new Integer(d.Value - int.Parse(s));

    public static Integer operator -(string s, Integer d) =>
        new Integer(int.Parse(s) - d.Value);

    public static implicit operator int(Integer d) =>
        d.Value;

    public static implicit operator Integer(int d) =>
        new Integer(d);
}

static void Main(string[] args)
{
    string x = "5";
    Integer y = 3;
    Integer result1 = x + y - y;
    Integer result2 = x - y + y;
    Console.WriteLine($"Result1: {result1.Value}\nResult2: {result2.Value}");
    Console.ReadKey();
}