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Alternative to out variables


What's the difference between the 'ref' and 'out' keywords?Parse JSON in C#Performance surprise with “as” and nullable typesIs there a reason for C#'s reuse of the variable in a foreach?Does anyone have benchmarks (code & results) comparing performance of Android apps written in Xamarin C# and Java?Why not inherit from List<T>?How to upgrade a project to C# 6Why can I assign 0.0 to enumeration values, but not 1.0How to use C# 6 with Web Site project type?TryParse with out var param













1















I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.










share|improve this question

















  • 5





    Just declare the variable before using it in TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);

    – Paweł Łukasik
    Mar 8 at 7:08






  • 1





    Got it, thanks.

    – Hose
    Mar 8 at 8:10















1















I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.










share|improve this question

















  • 5





    Just declare the variable before using it in TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);

    – Paweł Łukasik
    Mar 8 at 7:08






  • 1





    Got it, thanks.

    – Hose
    Mar 8 at 8:10













1












1








1








I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.










share|improve this question














I need to include double.TryParse(wordConf, out double wordConfDouble); in a script, but I get a feature out variable declaration is not available in c# 6 error message. Searching for it on Google, I can only see solutions for upgrading to C# 7 (which I am not allowed to do so in this project) so I wonder if someone could help me write an equivalent to this line that would work in any C# compiler.







c#






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 8 at 7:07









HoseHose

626




626







  • 5





    Just declare the variable before using it in TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);

    – Paweł Łukasik
    Mar 8 at 7:08






  • 1





    Got it, thanks.

    – Hose
    Mar 8 at 8:10












  • 5





    Just declare the variable before using it in TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);

    – Paweł Łukasik
    Mar 8 at 7:08






  • 1





    Got it, thanks.

    – Hose
    Mar 8 at 8:10







5




5





Just declare the variable before using it in TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);

– Paweł Łukasik
Mar 8 at 7:08





Just declare the variable before using it in TryParse. double wordConfDouble; double.TryParse(wordConf, out wordConfDouble);

– Paweł Łukasik
Mar 8 at 7:08




1




1





Got it, thanks.

– Hose
Mar 8 at 8:10





Got it, thanks.

– Hose
Mar 8 at 8:10












2 Answers
2






active

oldest

votes


















4














You don't need to inline declare a type for out-parameters.



Replace:



double.TryParse(wordConf, out double wordConfDouble);


With:



double wordConfDouble;
double.TryParse(wordConf, out wordConfDouble);





share|improve this answer






























    3














    It's just the inline declaration which is not supported in < C#7.0. Change your code to



    double wordConfDouble;
    double.TryParse(wordConf, out wordConfDouble);


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






    share|improve this answer


















    • 2





      Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.

      – Hose
      Mar 8 at 8:10










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    You don't need to inline declare a type for out-parameters.



    Replace:



    double.TryParse(wordConf, out double wordConfDouble);


    With:



    double wordConfDouble;
    double.TryParse(wordConf, out wordConfDouble);





    share|improve this answer



























      4














      You don't need to inline declare a type for out-parameters.



      Replace:



      double.TryParse(wordConf, out double wordConfDouble);


      With:



      double wordConfDouble;
      double.TryParse(wordConf, out wordConfDouble);





      share|improve this answer

























        4












        4








        4







        You don't need to inline declare a type for out-parameters.



        Replace:



        double.TryParse(wordConf, out double wordConfDouble);


        With:



        double wordConfDouble;
        double.TryParse(wordConf, out wordConfDouble);





        share|improve this answer













        You don't need to inline declare a type for out-parameters.



        Replace:



        double.TryParse(wordConf, out double wordConfDouble);


        With:



        double wordConfDouble;
        double.TryParse(wordConf, out wordConfDouble);






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 8 at 7:09









        Eric McLachlanEric McLachlan

        10817




        10817























            3














            It's just the inline declaration which is not supported in < C#7.0. Change your code to



            double wordConfDouble;
            double.TryParse(wordConf, out wordConfDouble);


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






            share|improve this answer


















            • 2





              Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.

              – Hose
              Mar 8 at 8:10















            3














            It's just the inline declaration which is not supported in < C#7.0. Change your code to



            double wordConfDouble;
            double.TryParse(wordConf, out wordConfDouble);


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






            share|improve this answer


















            • 2





              Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.

              – Hose
              Mar 8 at 8:10













            3












            3








            3







            It's just the inline declaration which is not supported in < C#7.0. Change your code to



            double wordConfDouble;
            double.TryParse(wordConf, out wordConfDouble);


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






            share|improve this answer













            It's just the inline declaration which is not supported in < C#7.0. Change your code to



            double wordConfDouble;
            double.TryParse(wordConf, out wordConfDouble);


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







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 8 at 7:09









            fubofubo

            30.8k969107




            30.8k969107







            • 2





              Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.

              – Hose
              Mar 8 at 8:10












            • 2





              Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.

              – Hose
              Mar 8 at 8:10







            2




            2





            Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.

            – Hose
            Mar 8 at 8:10





            Thank you for the answer and the reference. I had to mark the previous post as answer though, as it was posted just beforehand, but I do appreciate your reply.

            – Hose
            Mar 8 at 8:10

















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