int.TryParse syntatic sugar

Question

int.TryPrase is great and all, but there is only one problem...it takes at least two lines of code to use:

int intValue;
string stringValue = \"         


        

Answer 1

Because it essentially returns two values (success and the value), we really do need the two lines.

You could try a wrapper class, ie:

void Main()
{
    var result = simpleIntParser.TryParse("1");
    if(result)
    {
        Console.WriteLine((int)result);
    } else {
        Console.WriteLine("Failed");
    }

    result = simpleIntParser.TryParse("a");
    if(result)
    {
        Console.WriteLine((int)result);
    } else {
        Console.WriteLine("Failed");
    }


}

public class simpleIntParser
{
    public bool result {get; private set;}
    public int value {get; private set;}

    private simpleIntParser(bool result, int value)
    {
        this.result = result;
        this.value = value;
    }

    public static simpleIntParser TryParse(String strValue)
    {
        int value;
        var result = int.TryParse(strValue, out value);
        return new simpleIntParser(result, value);
    }

    public static implicit operator int(simpleIntParser m)
    {
        return m.value;
    }

    public static implicit operator bool(simpleIntParser m)
    {
        return m.result;
    }
}

It requires casting if the type is ambiguous (i.e. for Console.WriteLine()), but if you pass it as an integer parameter for example, no casting is required

Answer 2

This answer is only for those who use at least C# 7.

You can now declare the out parameter inline.

int.TryParse("123", out var result);

Exemplary usage:

if (int.TryParse("123", out var result)) {
    //do something with the successfully parsed integer
    Console.WriteLine(result);
} else {
    Console.WriteLine("That wasn't an integer!");
}

MSDN: https://docs.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-7#out-variables

Answer 3

I don't think there is anything really beautiful, but if you like this you get it down to one row:

string stringValue = "123"
int intValue = int.TryParse(stringValue, out intValue) ? intValue : 0;

Answer 4

One last addition to this NINE year-old question :). Bool parsing is a little different because if parsing fails, you don't want to return a default value, you want to return a NULL. This line does this (as of C# 7, I think):

return bool.TryParse(value, out bool result) ? (bool?)result : null;

That cast of the result is necessary, otherwise it cannot reconcile the differing types of the two return values.

Answer 5

int intValue = int.TryParse(stringValue, out intValue) ? intValue : 0;

Answer 6

Check out the StringExtensions class. It contains an AsInt(String,Int32) extension method that will attempt to convert a string and if unsuccessful populate it with the supplied Int32 value as default.

Example:

var intValue = "123".AsInt(-1);