C#•11mo ago

✅ How the "out" keyword works internally

Hello guys, sorry to disturb you all, I was just reading a bit about the

out

out

keyword. From what I have understood, the

out

out

keyword stores the "reference" of a variable tied to it. Any change method to that variable will reflect back when used later on.

I was wondering if someone can explain how the "out" keyword works internally related to what do we mean by "references", how are they stored in memory etc pls.

I have an overview of what happen, but I would really appreciate if someone can explain what's happening under the hood with the internal memory structure. (I think it's the same thing when we use the ref keyword, we modify addresses directly internally)

canton7•1/27/25, 1:14 PM

Do you know what a (C-style) pointer is?

FakerOP•1/27/25, 1:15 PM

yeah, I know a bit

FakerOP•1/27/25, 1:15 PM

I need to refresh but I know a bit how it works

canton7•1/27/25, 1:16 PM

A reference is the name that C# gives to a pointer (close enough: a reference is a pointer which the GC knows about)

canton7•1/27/25, 1:16 PM

So when you use

ref

ref

out

out

, instead of passing a value directly, the compiler passes a pointer, and the pointer contains the memory address that the value is stored at

FakerOP•1/27/25, 1:17 PM

yeah I see, but compared to C language where we would need to dereference the pointer etc, here we work directly with the variable

FakerOP•1/27/25, 1:17 PM

how is that possible

FakerOP•1/27/25, 1:17 PM

like it seems behind the scenes, everything is being done

canton7•1/27/25, 1:17 PM

The compiler knows that it's a pointer, and does the derefencing for you

FakerOP•1/27/25, 1:18 PM

canton7•1/27/25, 1:18 PM

Other languages do the same. See e.g. C++ and

references, or rust and auto-deref

FakerOP•1/27/25, 1:18 PM

yep I see

FakerOP•1/27/25, 1:20 PM

so for e.g, if I have something like out int resultout int result, what's happening is we are storing a reference to result, this allows us to modify the content of result directly and not a local copy of it. Now, when we write something like result = 20result = 20, behind the scenes, the compiler knows when to dereference things etc ? (was wondering, in which case would we need to dereference things here, like I was thinking when we write result = 20result = 20; behind the scenes, dereferencing occurs, but not sure)

canton7•1/27/25, 1:21 PM

(In C, you need to disambiguate between *pFoo = newValue*pFoo = newValue and pFoo = newValuepFoo = newValue (assigning a new value to the memory pointed to by the pointer, vs assigning a new memory address to the pointer itself). C# uses a different syntax for that. C# also doesn't let you have a reference which points to another reference, which simplifies the syntax)

Ccanton7 (In C, you need to disambiguate between `*pFoo = newValue` and `pFoo = newValue`...

FakerOP•1/27/25, 1:22 PM

ahh I see, here *pFoo = newValue*pFoo = newValue , we are assigning a new value to where the pointer points to?

FFaker so for e.g, if I have something like `out int result`, what's happening is we ar...

canton7•1/27/25, 1:22 PM

Yep, result = 20result = 20 means the compiler will look at the value of

result

result

, go to that memory address (an operation known as dereferencing), and write the value 20

FFaker ahh I see, here `*pFoo = newValue` , we are assigning a new value to where the p...

canton7•1/27/25, 1:22 PM

In C/C++, yes

FakerOP•1/27/25, 1:23 PM

yeah I see

FakerOP•1/27/25, 1:23 PM

thanks !

canton7•1/27/25, 1:24 PM

In C#, you can write things like:

int foo = 20;
int bar = 30;

ref int reference = ref foo;
reference = ref bar;

int foo = 20;
int bar = 30;

ref int reference = ref foo;
reference = ref bar;

Ccanton7 Yep, `result = 20` means the compiler will look at the value of `result`, go to ...

FakerOP•1/27/25, 1:24 PM

behind the scenes, result is a reference, when we assign 20 to result, it's like we are changing the value of the memory address at which result is, to 20, right ?

FFaker behind the scenes, result is a reference, when we assign 20 to result, it's like...

canton7•1/27/25, 1:25 PM

We're taking the value of

result

result

(which is a memory address), going to the memory with that address, and writing 20, yes

FakerOP•1/27/25, 1:25 PM

yeahh, thanks !!!

Ccanton7 In C#, you can write things like: ```cs int foo = 20; int bar = 30; ref int re...

FakerOP•1/27/25, 1:26 PM

will have to look at that later on, I've not yet study the ref keyword in methods

FFaker will have to look at that later on, I've not yet study the ref keyword in method...

canton7•1/27/25, 1:26 PM

It's entirely analogous to a

ref

ref

parameter passed to a method

Ccanton7 In C#, you can write things like: ```cs int foo = 20; int bar = 30; ref int re...

FakerOP•1/27/25, 1:30 PM

here, is it correct to say that we declare 2 variables of type int.

Then, we declare a reference that "points to" same memory address as foo, we use the int keyword because foo is an int?

Then later on, we change that reference to now points to where the reference of bar is.

Any change made to bar should reflect in the variable reference or vice-versa

FFaker here, is it correct to say that we declare 2 variables of type int. Then, we de...

canton7•1/27/25, 1:33 PM

ref int referenceref int reference declares a variable called referencereference whose type is a reference to an int

canton7•1/27/25, 1:33 PM

Yep for the other stuff

Ccanton7 `ref int reference` declares a variable called `reference` whose type is a refer...

FakerOP•1/27/25, 1:33 PM

ah ok, like we can say the reference is a memory address, "reference to an int" means, that memory address contains an int ?

canton7•1/27/25, 1:36 PM

It means that referencereference contains a memory address (rather than an int), and that value at that memory address is an int

FakerOP•1/27/25, 1:36 PM

yeppp, thanks, my doubts have been cleared, will be back if I have other questions but must be good for now, really appreciate

canton7•1/27/25, 1:36 PM

Great!

✅ How the "out" keyword works internally

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