Return empty array instead of null

Question

I created Settings class which I use to edit my application by .ini file. My Settings.ini file looks like this:

[ACCOUNT]
login=xyz
password=xyz
locations=1,2,5,8

Now I am getting theese values like this:

class Settings {
    public static IniFile Config = new IniFile(Directory.GetCurrentDirectory() + @"\Settings.ini");
    public static string Login { get { return Config.Read("login", "ACCOUNT"); } set { Config.Write("login", "ACCOUNT"); } }
    public static string Password { get { return Config.Read("password", "ACCOUNT"); } set { Config.Write("password", "ACCOUNT"); } }
    public static int[] Locations { get { return Array.ConvertAll(Config.Read("locations", "ACCOUNT").Split(','), s => int.Parse(s)); } set { Config.Write("locations", "ACCOUNT"); } }
}

The problem is, when my Settings.ini file has empty locations:

locations=

My variable Settings.Locations return null instead of empty array. I've tried doing something like this:

public static int[] Locations 
{ 
    get { return new int[] {Array.ConvertAll(Config.Read("locations", "ACCOUNT").Split(','), s => int.Parse(s))}; } 
    set { Config.Write("locations", "ACCOUNT"); } 
}

But that is just not working. I cant convert int[] to int. Do you have any ideas how can I return empty array?

I dont get it, Config.Read("locations", "ACCOUNT") returns null if there are no locations? Because your Locations property never returns null because Array.ConvertAll never returns null. But if Config.Read returns null string.Split throws an exception. — Tim Schmelter
– Tim Schmelter, Commented May 4, 2017 at 11:56
Instead of Array.ConvertAll you should use a loop and int.TryParse to try-parse every token. Then you can handle invalid input — Tim Schmelter
– Tim Schmelter, Commented May 4, 2017 at 12:14

Tao Gómez Gil · Accepted Answer · 2017-05-04 11:47:28Z

24

You can do it explicitly like this:

public static int[] Locations
{
    get
    {
        string locations = Config.Read("locations", "ACCOUNT");
        if (locations == null)
        {
            return new int[0];
        }
        return locations
                .Split(',')         // split the locations separated by a comma
                .Select(int.Parse)  // transform each string into the corresponding integer
                .ToArray();         // put the resulting sequence (IEnumerable<int>) into an array of integers
    }
    set
    {
        Config.Write("locations", "ACCOUNT");
    }
}

answered May 4, 2017 at 11:47

Tao Gómez Gil

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2 Comments

Brak Danych Over a year ago

I had to change to: if (string.IsNullOrEmpty(works)) { and it is working like a charm. Thanks :)

Mario Lopez Over a year ago

VS complaints about that now. It suggests Array.Empty<int>();.

rory.ap · Accepted Answer · 2017-05-04 11:41:34Z

4

First of all, you're jamming too much into one line so it makes it really hard to read let alone troubleshoot. What you need is something like this:

public static int[] Locations 
{ 
    get 
    { 
        int[] values = Array.ConvertAll(Config.Read("locations", "ACCOUNT").Split(','), 
            s => int.Parse(s)) ?? new int[] { };
        return values; 
    } 
    set 
    { 
        Config.Write("locations", "ACCOUNT"); 
    } 
}

Notice I have added ?? new int[] { } to the end of the first statement, which is called the null coalescing operator. It will return an empty array if the other array is null.

It's a matter of preference, but the reason I separated the getter into two lines is so I can debug and break before it returns to observe the return value. You could also break on the last bracket instead and observe the return value in the Locals window.

edited May 4, 2017 at 11:41

answered May 4, 2017 at 11:38

rory.ap

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2 Comments

Camilo Terevinto Over a year ago

Why declare values and not use it at all? Besides of the missing ;

Brak Danych Over a year ago

I was getting Format exception and I didint noticed it. Your code did not work aswell.

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