TryParse create inline parameter?

Question

Is there any way in C# to create a variable inline? Something like this:

int x = int.TryParse("5", out new int intOutParameter) ? intOutParameter : 0;

Don´t you think that this is more useful than creating a variable outside and then never use it again?

Answer 1

That syntax – called declaration expressions – was on the proposed feature list for the next version of C# (version 6).

You're not the only one to think it is useful. For instance making a complete TryParse call an expression (no need for a statement to declare the variable).

However it has been dropped from the ongoing work to C#6.

I'm sure I'm not the only one hoping it will make a return in a future version.It is included in C#7 as a declaration (no need for new):

int x = int.TryParse("5", out int intOutParameter) ? intOutParameter : 0;

Answer 2

Inline declarations for out params is a new suggested feature in C# that might be standard one day, see e.g. Probable C# 6.0 features illustrated, section 9. The expected/proposed syntax:

int.TryParse("5", out int x); // this declares (and assigns) a new variable x

Edit: This out variable syntax was eventually included in C# 7.0 (Visual Studio 2017); you can also use out var x.

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Addition: People come up with fun extension methods. I tried to make a generic one:

public delegate bool TryParser<TResult>(string s, out TResult result);

public static class FunExtensions
{
  public static T TryParse<T>(this string str, TryParser<T> tryParser)
  {
    T outResult;
    tryParser(str, out outResult);
    return outResult;
  }
}

This can be used like this:

  var x = "5".TryParse<int>(int.TryParse);
  var y = "01/01".TryParse<DateTime>(DateTime.TryParse);
  var z = "bad".TryParse<decimal>(decimal.TryParse);

and so on. I was hoping the compiler would infer T from usage, so that one could say simply:

  var x = "5".TryParse(int.TryParse);  // won't compile

but it appears you have to explicitly specify the type argument to the method.