Is it Out can be used for passing a reference type (i.e) Objects?

Question

Consider the following example, here int i is passed for the reference.

My question is, Can i pass a reference type to out? like object (i.e) static void sum(out OutExample oe)

class OutExample
{
   static void Sum(out int i)
   {
      i = 5;
   }
   static void Main(String[] args)
   {
     int val;
     Sum(out val);
     Console.WriteLine(val);
     Console.Read();
   }
}

Now the below code having some error,

class OutExample
{
  int a;
   static void Sum(out OutExample oe)
   {
    oe.a = 5;
   }
   static void Main(String[] args)
   {
    int b;
    OutExample oe1=new OutExample();
    Sum(out oe);
    oe.b=null;
    Console.WriteLine(oe.b);
    Console.Read();
   }
   }

Finally Got the Answer!

class OutExample
{
int a;
int b;

static void Sum(out OutExample oe)
 {
   oe = new OutExample();
   oe.a = 5;
 }

static void Main(String[] args)
{ 
   OutExample oe = null; 
   Sum(out oe);
   oe.b = 10;
   Console.WriteLine(oe.a);
   Console.WriteLine(oe.b);
   Console.Read();
}
}

@Balaji so... did you try `static void sum(out OutExample oe)` ? what happened? — Marc Gravell, Dec 18 '14 at 11:12
ya, but it is showing some error! The out parameter 'oe' must be assigned to before control leaves the current method — Balaji, Dec 18 '14 at 11:17
@Balaji that happens for *all* `out` parameters if you don't assign a value to it; if you remove the `i=5`, the same will happen in the code in your question. — Marc Gravell, Dec 18 '14 at 11:19

score 1 · Answer 1 · answered Dec 18 '14 at 11:19

1

Yes...

static void Sum(out OutExample oe)
{
    oe = null;
    // or: oe = new OutExample();
}
class OutExample {}

answered Dec 18 '14 at 11:19

Marc Gravell

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If it is **null** how can i use it in **Main()**? – Balaji Dec 18 '14 at 11:25
@Balaji that is up to `Main`! and of course, `Sum` could set it to a non-null instance – Marc Gravell Dec 18 '14 at 11:34
can u please fully edit the above pgm and your answer too? – Balaji Dec 18 '14 at 11:38

t3chb0t · Accepted Answer · 2014-12-18T11:32:30.337

1

You must create a new OutExample inside the Sum method:

class OutExample
{
   int a;
   int b;

   static void Sum(out OutExample oe)
   {
       oe = new OutExample();
       oe.a = 5;
   }

   static void Main(String[] args)
   { 
       OutExample oe = null; 
       Sum(out oe);
       oe.b = 10;
       Console.WriteLine(oe.a);
       Console.WriteLine(oe.b);
       Console.Read();
   }
}

edited Dec 18 '14 at 11:32

answered Dec 18 '14 at 11:24

t3chb0t

16,340
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Ok But the **b** is underlined red of telling, Class doesn't contain a definition for **b** – Balaji Dec 18 '14 at 11:30
This is true. You defined `b` inside the `Main` scope so it's a local variable and not a class field like the `a`. – t3chb0t Dec 18 '14 at 11:31
the definition of **b** is within the **Main** only, then how can it be a non local variable? – Balaji Dec 18 '14 at 11:34
It is a local variable but only in the scope of the `Main` method and not the entire class. I modified the example – t3chb0t Dec 18 '14 at 11:36
Fantastic man, its working well... – Balaji Dec 18 '14 at 11:39

score 1 · Answer 3 · answered Dec 18 '14 at 11:40

1

I would suggest you to rethink something. Reference type are a reference to a storage location. Passing it in out you are passing a reference to this references. Why don't you pass directly by ref?

answered Dec 18 '14 at 11:40

faby

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I think @Balaji is using `out` because he wanted to learn how it works. Questions whether it's a good or a bad practice will come later :-) – t3chb0t Dec 18 '14 at 11:50

Is it Out can be used for passing a reference type (i.e) Objects?

3 Answers3