C#•3y ago

✅ reference type values are strange..

static void bar(ref int[] baz, int newElem)
{
    int newLength = baz.Length + 1;
    int [] quux = new int [newLength];

    for(int i = 0; i < baz.Length; i++)
        quux[i] = baz[i];

    quux[newLength - 1] = newElem;

    baz = quux;
}
    
static void Main() 
{
    int[] foo = {1, 2, 3, 4};
    foreach(int i in foo)
        Console.Write(i + " ");
    
    Console.WriteLine();
    bar(ref foo, 42);
    
    foreach(int i in foo)
        Console.Write(i + " ");
}

static void bar(ref int[] baz, int newElem)
{
    int newLength = baz.Length + 1;
    int [] quux = new int [newLength];

    for(int i = 0; i < baz.Length; i++)
        quux[i] = baz[i];

    quux[newLength - 1] = newElem;

    baz = quux;
}
    
static void Main() 
{
    int[] foo = {1, 2, 3, 4};
    foreach(int i in foo)
        Console.Write(i + " ");
    
    Console.WriteLine();
    bar(ref foo, 42);
    
    foreach(int i in foo)
        Console.Write(i + " ");
}

63 Replies

wcasaOP•3y ago

why do i have to pass by reference array that I am going to change? arrays are reference type value, what is the point of passing them by reference?

cap5lut•3y ago

to make baz = quux; work

void Example1(int[] array)
{
  array = new int[] { 1, 2 };
}
void Example2(ref int[] array)
{
  array = new int[] { 1, 2 };
}

static void Main()
{
  var data = new int[0];
  Example1(data);
  foreach (var e in data) Console.WriteLine(e); // will not print anything because data itself wasnt modified

  Example2(ref data);
  foreach (var e in data) Console.WriteLine(e); // will not print 1 and 2 because data itself was modified

}

void Example1(int[] array)
{
  array = new int[] { 1, 2 };
}
void Example2(ref int[] array)
{
  array = new int[] { 1, 2 };
}

static void Main()
{
  var data = new int[0];
  Example1(data);
  foreach (var e in data) Console.WriteLine(e); // will not print anything because data itself wasnt modified

  Example2(ref data);
  foreach (var e in data) Console.WriteLine(e); // will not print 1 and 2 because data itself was modified

}

wcasaOP•3y ago

you mean will print 1 and 2 because data itself was modified?

cap5lut•3y ago

yes

wcasaOP•3y ago

(last comment) but why isnt it modified

cap5lut•3y ago

by data i mean the local variable in Main() was modified, because it is a reference (like a pointer) to it

wcasaOP•3y ago

shouldnt method without reference point to the same memory as array is in

cap5lut•3y ago

in Example1 array is just a copy of the reference stored in data

wcasaOP•3y ago

ooooohhhh

cap5lut•3y ago

yeah, but with baz = quux u basically let it point to other memory

wcasaOP•3y ago

but how does it make difference

wcasaOP•3y ago

wcasaOP•3y ago

imagine third box as copy of reference that was created by method

cap5lut•3y ago

hmm, lets take it with int instead of int[]:

void Example1(int x) { x = x + 1; }
void Example2(ref int x) { x = x + 1 }
void Main()
{
  int value = 0;
  Example1(value);
  Example2(ref value);
}

void Example1(int x) { x = x + 1; }
void Example2(ref int x) { x = x + 1 }
void Main()
{
  int value = 0;
  Example1(value);
  Example2(ref value);
}

wcasaOP•3y ago

but int is value type, not reference type this one will change

cap5lut•3y ago

but a reference itself is something similar to a value type when Example1 will be called, another stack frame will be created, which introduces the parameter x, then the value of value will be copied to x (in this case 0), then the method will be executed and x will be 1, then the method will be left and the stack frame will be discarded (x and its modification is lost)

wcasaOP•3y ago

yes

cap5lut•3y ago

when Example2 will be called its a bit different: a stack frame will be created, which introduces a pointer to an int will be introduced (thats more or less x) then the pointer of value (from Main()) will be copied to that x the the method will execute and the memory x is pointing to will be modified (thats value on the stack frame of Main()) the method will be left, the stack frame is gone and thus the copy of the pointer of value, but its modification not, because its on the stack frame of Main() in c it would look like

void Example1(int x) { x = x + 1; }
void Example2(int* x) { *x = *x + 1 }
void Main()
{
  int value = 0;
  Example1(value);
  Example2(&value);
}

void Example1(int x) { x = x + 1; }
void Example2(int* x) { *x = *x + 1 }
void Main()
{
  int value = 0;
  Example1(value);
  Example2(&value);
}

if thats more understandable for u

wcasaOP•3y ago

yeah, i totally understand this but its for the value type

cap5lut•3y ago

so now comming back for reference types:

Aaron•3y ago

ref int[] isn't a reference to the array itself

cap5lut•3y ago

int[] value; doesnt store the array, but a pointer to a memory block on the heap

Aaron•3y ago

it's a reference to the local variable in your method

cap5lut•3y ago

if u would just modify the elements of the array, it wouldnt matter if its a int[] or ref int[] parameter, because u just modify the memory on the heap. but as u want to assign it to a new memory block baz = quux; it makes a difference if u just modify the stack frame's local variable or the pointer/reference itself

wcasaOP•3y ago

oohhhh missed the fact that quux is new memory block

cap5lut•3y ago

;p so, is ur question answered? if yes, please $close the thread

MODiX•3y ago

Use the /close command to mark a forum thread as answered

wcasaOP•3y ago

wait a sec, thinking it over wait fucking what so its like referencing to a reference?

Aaron•3y ago

yes that's exactly what it is a ref int[] is effectively a int** + a length

333fred•3y ago

Whenever you see a reference type variable in code, you can think of it as a pointer So string s is a pointer to character data

cap5lut•3y ago

that why i took the int example first, to verify the general understanding ;p

333fred•3y ago

int[] is a pointer to int data

wcasaOP•3y ago

wow

333fred•3y ago

Adding a ref modifier is adding a pointer to the signature So ref int[] is effectively int** But safer

wcasaOP•3y ago

so i took reference, and dereferenced it to a new chank of memory that i have created?

333fred•3y ago

In which example?

cap5lut•3y ago

yes

wcasaOP•3y ago

static void bar(ref int[] baz, int newElem)
{
    int newLength = baz.Length + 1;
    int [] quux = new int [newLength];

    for(int i = 0; i < baz.Length; i++)
        quux[i] = baz[i];

    quux[newLength - 1] = newElem;

    baz = quux;
}
    
static void Main() 
{
    int[] foo = {1, 2, 3, 4};
    foreach(int i in foo)
        Console.Write(i + " ");
    
    Console.WriteLine();
    bar(ref foo, 42);
    
    foreach(int i in foo)
        Console.Write(i + " ");
}

static void bar(ref int[] baz, int newElem)
{
    int newLength = baz.Length + 1;
    int [] quux = new int [newLength];

    for(int i = 0; i < baz.Length; i++)
        quux[i] = baz[i];

    quux[newLength - 1] = newElem;

    baz = quux;
}
    
static void Main() 
{
    int[] foo = {1, 2, 3, 4};
    foreach(int i in foo)
        Console.Write(i + " ");
    
    Console.WriteLine();
    bar(ref foo, 42);
    
    foreach(int i in foo)
        Console.Write(i + " ");
}

333fred•3y ago

Right, so bar has an effectively int** parameter

wcasaOP•3y ago

wow wowowowowoowowow and all of the arrays, classes are all same as just pointer to smth? like int*?

cap5lut•3y ago

all reference types are

wcasaOP•3y ago

damn c# is cool

cap5lut•3y ago

effectively they work like pointers, but because this is managed code its a bit different because the memory manager could move the memory blocks around, thus pointers would be invalid if they would contain "real" addresses

Aaron•3y ago

they do contain the real addresses

wcasaOP•3y ago

thank you everyone, thats a big help! !close

Accord•3y ago

Closed!

cap5lut•3y ago

yeah but there is more to it because of the memory management i mean, or am i wrong here? basically what i meant is, that the address can change without every touching that one reference

Aaron•3y ago

yeah, the GC can move things around

wcasaOP•3y ago

btw can i somehow pass the copy of array itself?

cap5lut•3y ago

copy the array, and pass the copy ;p

333fred•3y ago

Not in one easy action

wcasaOP•3y ago

static void bar(int[] baz)
{
    int [] quux = new int[baz.Length];

    for(int i = 0; i < baz.Length; i++)
        quux[i] = baz[i];

    //here you go quux is now copy
    
}

static void bar(int[] baz)
{
    int [] quux = new int[baz.Length];

    for(int i = 0; i < baz.Length; i++)
        quux[i] = baz[i];

    //here you go quux is now copy
    
}

is this ok?

cap5lut•3y ago

isnt it just for example new int[] { 1, 2, 3 }.ToArray()?

333fred•3y ago

Sure.

Aaron•3y ago

fastest way is just array.AsSpan().ToArray() I think

333fred•3y ago

Probably

Aaron•3y ago

but all of them work fine

333fred•3y ago

Well, it should be pretty close in perf to what was written above, but it's certainly easier to type

wcasaOP•3y ago

ok, thank in advance

Aaron•3y ago

just doing .ToArray() on the array gets you the LINQ method, which has a fast path for lists and the like where length is known, but also has to do a type check and multiple vcalls to get there so ¯\_(ツ)_/¯

cap5lut•3y ago

well, for optimization just throw a benchmark on all, but i guess the span version would be the fastest

333fred•3y ago

I meant in comparison to the foreach

Aaron•3y ago

yeah I know, was just comparing to caps answer

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✅ reference type values are strange..

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