Out argument of function call should maintain initialized value

Question

I have defined a function where one of the parameters is out. Here, when the function call is made, I pass either a an initialized or an uninitialized argument. Now, in the case of an initialized argument, how do I make the callee not change the value of the out parameter?

I cant use a ref here because sometimes I do send an uninitialized argument. Ex:

void fun1()
{

    int x = 3;
    fun2 (out x);
    int y;
    fun2(out y);

}

void fun2(out int x)
{

    ...

}

Here, I dont want the x to lose the value 3 once control goes to fun2.

If you put some code to show what you are doing, it would help us visualize how to help you. — Guga, Oct 03 '13 at 21:06
It's a little hard to understand what you trying to achieve. "out" means that the initialization (and value) is the responsibility of the called method. You are asking for the complete opposite. — Tormod, Oct 03 '13 at 21:09
In the example above, I just want to ensure that if the out argument is already initialized in fun1 before fun2 call, it should keep that value intact else, it can initialize the argument in fun2. But, since it is out, fun2 has to assign a value to the parameter because of which in the first call, the value 3 is lost.\ — Tyler Durden, Oct 03 '13 at 21:12
Just out of curiosity, did you know that int is always initialized to 0? — Jesko R., Oct 03 '13 at 21:18

score 3 · Accepted Answer · answered Oct 03 '13 at 21:10

3

From out C# - MSDN

Although variables passed as out arguments do not have to be initialized before being passed, the called method is required to assign a value before the method returns.

Since a value must be assigned to the parameter with out, you can't save the value in function. It would be better if you make a copy of the variable before calling the function. like:

int x = 1;
int backupX = x;
fun2(out x);

answered Oct 03 '13 at 21:10

Habib

219,104
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436

That sounds right. Thanks. Which brings me to my next question. What is the difference between ref? and out. – Tyler Durden Oct 03 '13 at 21:14
@TylerDurden, see this question. It has very detailed answer See this question: http://stackoverflow.com/questions/1516876/when-to-use-ref-vs-out – Habib Oct 03 '13 at 21:15
I did see that question earlier but it doesnt explain the difference between nullable ref and out – Tyler Durden Oct 03 '13 at 21:17
@TylerDurden, what do you mean by nullable ? `Nullable` or un-initialized `int` ? – Habib Oct 03 '13 at 21:18
I mean Nullable. Uninitialized int doesnt exist right because it is by default 0 right? – Tyler Durden Oct 03 '13 at 21:22
if you have `int x;` that is **unassigned** `int`. – Habib Oct 03 '13 at 21:24

score 0 · Answer 2 · answered Oct 03 '13 at 21:23

0

Maybe I get all wrong, but this sounds like you just want to define a method like this

void caller(){
int x=5;
int y = doSomething(x);
}

int doSomething(int x){
 return x+1;
}

or in case you want a null state use:

void caller(){
int? x=5;
int y = doSomething(x);
}

int doSomething(int? x){
 if (x == null)
    return x;
 return x+1;
}

answered Oct 03 '13 at 21:23

Jesko R.

827
10
21

this is fine only if there is one argument. What if there are more? Which is why I am using ref/out and making the function return type as void – Tyler Durden Oct 03 '13 at 21:27
define classes and use objects in such case – Jesko R. Oct 03 '13 at 21:29

Out argument of function call should maintain initialized value

2 Answers2