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Home/ Questions/Q 292365
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T06:10:25+00:00 2026-05-12T06:10:25+00:00

if I run this C# code int realInt = 3; string foo = bar;

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if I run this C# code

int realInt = 3;  
string foo = "bar";  
Int32.TryParse(foo, out realInt); 

Console.WriteLine(realInt);  
Console.Read();

I get 0. And I would like to know why! Cause I cannot find any reason why it would. This forces me to make temp variables for every parsing. So please! Great coders of the universe, enlighten me!

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  1. Editorial Team
    Editorial Team
    2026-05-12T06:10:25+00:00Added an answer on May 12, 2026 at 6:10 am

    It is “out”, not “ref”. Inside the method it has to assign it (without reading it first) to satisfy the meaning of “out”.

    Actually, “out” is a language concern (not a framework one) – so a managed C++ implementation could probably ignore this… but it is more consistent to follow it.

    In reality; if the method returns false you simply shouldn’t look at the value; treat it as garbage until it is next assigned. It is stated to return 0, but that is rarely useful.


    Also – if it didn’t do this (i.e. if it preserved the value); what would this print:

    int i;
    int.TryParse("gibber", out i);
    Console.WriteLine(i);
    

    That is perfectly valid C#… so what does it print?

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