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Reference type still needs pass by ref?
(6 answers)
Closed 9 years ago.
class Test
{
static void Func(StringBuilder myString)
{
myString.Append ("test");
myString = null;
}
static void Main()
{
StringBuilder s1 = new StringBuilder();
Func(s1);
Console.WriteLine (s1);
}
}
The output is "Test", why isn't it null?
If s1 is passed by reference to Func(), then why does myString.Append("test") change it, but myString = null does not?
Thanks in advance.
You're passing the reference to the object by value.
I.e. the address of the object is passed by value, but the address to the object and the object is the same. So when you call your method, the VM copies the reference; you're just changing a copy.
On myString.Append you're using the copied reference to get to the object (still only one object) which is then changed/invoked.
What you think you're doing is this:
class Test
{
static void Func(ref StringBuilder myString)
{
myString.Append("test");
myString = null;
}
static void Main(string[] args)
{
StringBuilder s1 = new StringBuilder();
Func(ref s1);
Console.WriteLine(s1);
}
}
A longer explanation: http://javadude.com/articles/passbyvalue.htm
if s1 is passed by reference to Func(), then why does
myString.Append("test") does change it
Its not passed by references, instead its address value is passed to the function.
Suppose S1 is pointing to a memory location 0x48, This will be passed to Func, where Func parameter myString will start pointing to that location. Later when you Append a text to that location, it gets added. But later when you assign null to myString, it starts pointing to nothing, but the original location remains unchanged.
You should see: Parameter passing in C# by Jon Skeet
Consider the following diagrams.
In Step 1, you create a new object of type StringBuilder and let S1 refers to that object.
In Step 2, after passing S1 to Func, now myString also points to the same object in memory.
Later when you Append, text through myString, it updates the original object, since myString is pointing to the memory location held by the object.
In Step 3, when you assign null to the myString it looses the address for StringBuilder object, but this doesn't change anything in the object or the pointer S1.
In C#, the pointer (reference) to a complex type, by default, is passed by value. You must specify that it is passed by reference, like this:
static void Func(ref StringBuilder myString)
{
myString.Append ("test");
myString = null;
}
add the ref to the function to pass it by ref.
static void Func(ref StringBuilder myString)
{
myString.Append ("test");
myString = null;
}
Then call it like:
Func(ref s1);
Because you pass a reference to your StringBuilder s1 to Func.
Inside Func, the reference to that same StringBuilder instance is called "myString".
You first tell it to do something ("myString.Append("Test")), then set the reference INSIDE Func to null. The result of Append is visible to Main(); the reference being set to null does not influence s1 in Main.
Related
I am new to C# and I have been messing around with 'ref', 'out' and pointers, and I have a general question about how 'ref' works, especially when using objects and not primitive types. Say this is my method:
public void foo(ref Point p) {
p.set(1,1); // the x/y values are updated without constructing a new Point
}
and a similar method:
public void bar(Point p) {
p.set(1,1); // the x/y values are updated without constructing a new Point
}
EDIT: Point is a class in both cases
Both work, but is one more cost effective than the other? I know in C++ if you pass in a pointer you are only giving the memory address; from my understanding of C#, you cannot pass in an Object* into a method because of the automatic garbage collection. Does 'ref' pin an object to a location? Also, if you pass in an object to a method, like 'bar' above, is it passing a copy of the object or is it passing a pointer/reference?
Clarification: In my book I have, it does say if you want a method to update a primitive, such as int, you need to use ref (out if it is not initialized) or a *. I was asking if the same holds true for objects, and if passing an object as a parameter rather than a ref to an object costs more.
If your type is a struct, ref is roughly equivalent to a pointer to that struct. No new instances are created here. The difference (from passing it without ref) is that you can now mutate the original struct instance contained in that variable.
If your type is a class, ref simply adds one more level of indirection. No new instances are created here either. The difference (from passing it without ref) is that you can now entirely replace (not just mutate) the original class instance referenced by that variable with something else.
Since no new instances are created in either case, the garbage collector probably won't care about this in any important way.
In fact, class is a reference type, it mean that a variable of a reference type hold a reference to it's data instead of holding is data directly like value type.
When you pass a variable of a reference type as method parameter, it pass the reference to that data, not the data itself. So if update some properties of your object, the update is reflected in the original variable, except if you reassign the parameter.
Example from MSDN :
class PassingRefByVal
{
static void Change(int[] pArray)
{
pArray[0] = 888; // This change affects the original element.
pArray = new int[5] {-3, -1, -2, -3, -4}; // This change is local.
System.Console.WriteLine("Inside the method, the first element is: {0}", pArray[0]);
}
static void Main()
{
int[] arr = {1, 4, 5};
System.Console.WriteLine("Inside Main, before calling the method, the first element is: {0}", arr [0]);
Change(arr);
System.Console.WriteLine("Inside Main, after calling the method, the first element is: {0}", arr [0]);
}
}
/* Output:
Inside Main, before calling the method, the first element is: 1
Inside the method, the first element is: -3
Inside Main, after calling the method, the first element is: 888
*/
Passing a variable of a reference type with the ref keyword will reflect any change to the original variable, even if you reassin the parameter.
Example from MSDN :
class PassingRefByRef
{
static void Change(ref int[] pArray)
{
// Both of the following changes will affect the original variables:
pArray[0] = 888;
pArray = new int[5] {-3, -1, -2, -3, -4};
System.Console.WriteLine("Inside the method, the first element is: {0}", pArray[0]);
}
static void Main()
{
int[] arr = {1, 4, 5};
System.Console.WriteLine("Inside Main, before calling the method, the first element is: {0}", arr[0]);
Change(ref arr);
System.Console.WriteLine("Inside Main, after calling the method, the first element is: {0}", arr[0]);
}
}
/* Output:
Inside Main, before calling the method, the first element is: 1
Inside the method, the first element is: -3
Inside Main, after calling the method, the first element is: -3
*/
MSDN documentation.
Quick distinction between a class vs. a struct:
A class is a reference type. When an object of the class is created,
the variable to which the object is assigned holds only a reference to
that memory. When the object reference is assigned to a new variable,
the new variable refers to the original object. Changes made through
one variable are reflected in the other variable because they both
refer to the same data.
A struct is a value type. When a struct is
created, the variable to which the struct is assigned holds the
struct's actual data. When the struct is assigned to a new variable,
it is copied. The new variable and the original variable therefore
contain two separate copies of the same data. Changes made to one copy
do not affect the other copy.
https://msdn.microsoft.com/en-us/library/ms173109.aspx
Your example is tricky because in c# Point is an immutable struct, not an object.
Hopefully this example will help show what happens with structs and objects with and without ref.
public static void StructTest()
{
var fooStruct = new MyStruct();
var barStruct = new MyStruct();
Console.WriteLine(fooStruct.Value); // prints 0
Console.WriteLine(barStruct.Value); // prints 0
fooStruct(ref fooStruct);
barStruct(barStruct);
// Struct value only changes when passed by reference.
Console.WriteLine(fooStruct.Value); // prints 1
Console.WriteLine(barStruct.Value); // prints 0
}
public void fooStruct(ref MyStruct m)
{
m.Value++;
}
public void barStruct(MyStruct m)
{
m.Value++;
}
public static void ObjectTest()
{
var fooObject = new MyObject();
var barObject = new MyObject();
Console.WriteLine(fooObject.Value); // prints 0
Console.WriteLine(barObject.Value); // prints 0
fooObject(ref fooObject);
barObject(barObject);
// Objects are automatically passed by reference. No difference.
Console.WriteLine(fooObject.Value); // prints 1
Console.WriteLine(barObject.Value); // prints 1
fooSetObjectToNull(ref fooObject);
barSetObjectToNull(barObject);
// Reference is actually a pointer to the variable that holds a reference to the object.
Console.WriteLine(fooObject == null); // prints true
Console.WriteLine(barObject == null); // prints false
}
public void fooObject(ref MyObject m)
{
m.Value++;
}
public void barObject(ref MyObject m)
{
m.Value++;
}
public void fooSetObjectToNull(ref MyObject m)
{
m = null;
}
public void barSetObjectToNull(MyObject m)
{
m = null;
}
I understand that a string is sent by value even though it is a reference type, but if my string was tens of MB in size and i wanted to send it as an argument.
should i send it by reference or value?
private int GetIndexOfNext(string String,int SearchStartIndex,char TargetChar)
Or
private int GetIndexOfNext(ref string String,int SearchStartIndex,char TargetChar)
I understand that a string is sent by value even though it is a
reference type
For strings or other reference types, their address is passed by value. It is not the value which gets passed, so for your case it doesn't matter.
The reason parameter passing involving strings looks different is because strings are immutable (and that when you try to modify string's content).
You should see: Parameter passing in C# by Jon Skeet
It's fine to pass it as a string without the ref. The string isn't copied, you've just passed a copy of the reference to the method, rather than a copy of the string.
A small (unsafe) test will show that it's the string address that is passed and not a copy of the string.
static void Main(string[] args)
{
String s = "aaa";
Console.WriteLine(s); // Prints aaa
F(s);
Console.WriteLine(s); // Prints aba
Console.ReadLine();
}
static unsafe void F(String s)
{
fixed (char* p = s)
{
p[1] = 'b';
}
}
The string type in C# is a reference type, and passing a reference type argument by value copies the reference so that I don't need to use the ref modifier. However, I need to use the ref modifier for modifying the input string. Why is this?
using System;
class TestIt
{
static void Function(ref string input)
{
input = "modified";
}
static void Function2(int[] val) // Don't need ref for reference type
{
val[0] = 100;
}
static void Main()
{
string input = "original";
Console.WriteLine(input);
Function(ref input); // Need ref to modify the input
Console.WriteLine(input);
int[] val = new int[10];
val[0] = 1;
Function2(val);
Console.WriteLine(val[0]);
}
}
The reason you need to ref the string parameter is that even though you pass in a reference to a string object, assigning something else to the parameter will just replace the reference currently stored in the parameter variable. In other words, you have changed what the parameter refers to, but the original object is unchanged.
When you ref the parameter, you have told the function that the parameter is actually an alias for the passed-in variable, so assigning to it will have the desired effect.
EDIT: Note that while string is an immutable reference type, that's not too relevant here. Since you're just trying to assign a new object (in this case the string object "modified"), your approach wouldn't work with any reference type. For example, consider this slight modification to your code:
using System;
class TestIt
{
static void Function(ref string input)
{
input = "modified";
}
static void Function2(int[] val) // don't need ref for reference type
{
val = new int[10]; // Change: create and assign a new array to the parameter variable
val[0] = 100;
}
static void Main()
{
string input = "original";
Console.WriteLine(input);
Function(ref input); // need ref to modify the input
Console.WriteLine(input);
int[] val = new int[10];
val[0] = 1;
Function2(val);
Console.WriteLine(val[0]); // This line still prints 1, not 100!
}
}
Now, the array test "fails", because you're assigning a new object to the non-ref parameter variable.
It helps to compare string to a type that is like string but is mutable. Let's see a short example with StringBuilder:
public void Caller1()
{
var builder = new StringBuilder("input");
Console.WriteLine("Before: {0}", builder.ToString());
ChangeBuilder(builder);
Console.WriteLine("After: {0}", builder.ToString());
}
public void ChangeBuilder(StringBuilder builder)
{
builder.Clear();
builder.Append("output");
}
This produces:
Before: input
After: output
So we see that for a mutable type, i.e. a type that can have its value modified, it is possible to pass a reference to that type to a method like ChangeBuilder and not use ref or out and still have the value changed after we called it.
And notice that at no time did we actually set builder to a different value in ChangeBuilder.
By contrast, if we do the same thing with string:
public void Caller2()
{
var s = "input";
Console.WriteLine("Before: {0}", s);
TryToChangeString(s);
Console.WriteLine("After: {0}", s);
}
public void TryToChangeString(string s)
{
s = "output";
}
This produces:
Before: input
After: input
Why? Because in TryToChangeString we are not actually changing the contents of the string referenced by the variable s, we are replacing s with an entirely new string. Furthermore, s is a local variable to TryToChangeString and so replacing the value of s inside the function has no effect on the variable that was passed in to the function call.
Because a string is immutable, there is no way, without using ref or out, to affect the callers string.
Finally, the last example does what we want with string:
public void Caller3()
{
var s = "input";
Console.WriteLine("Before: {0}", s);
ChangeString(ref s);
Console.WriteLine("After: {0}", s);
}
public void ChangeString(ref string s)
{
s = "output";
}
This produces:
Before: input
After: output
The ref parameter actually makes the two s variables aliases for each other. It's as though they were the same variable.
Strings are immutable - you are not modifying the string but replacing the object the reference points to with another one.
Compare that with e.g., a List: To add Items, you don't need ref. To replace the entire list with a different object, you need ref (or out).
This is the case for all immutable types. string happens to be immutable.
In order to change the immutable type outside of the method, you must change the reference. Therefore either ref or out is required to have an effect outside of the method.
Note: It's worth noting that in your example, you are calling out a particular case that does not match the other example: you are actually pointing to a different reference rather than simply changing the existing reference. As noted by dlev (and the Skeet himself in my comments), if you did the same for all other types (e.g., val = new int[1]), including mutable ones, then you will "lose" your changes once the method returns because they did not happen to the same object in memory, unless you use ref or out like you did with string above.
To hopefully clarify:
You are passing in a pointer that points to your object in memory. Without ref or out, a new pointer is made that points to the exact same location, and all changes happen using the copied pointer. Using them, the same pointer is used and all changes made to the pointer are reflected outside the method.
If your object is mutable, then that means that it can be changed without creating a new instance of the object. If you create a new instance, then you must point to somewhere else in memory, which means you must change your pointer.
Now, if your object is immutable, then that means that it cannot be changed without creating a new instance.
In your example, you created a new instance of a string (equal to "modified") and then changed the pointer (input) to point to that new instance. For the int array, you changed one of the 10 values effectively pointed to by val, which does not require messing with val's pointer--it simply goes to where you want (the first element of the array), and then modifies that first value, in-place.
A more similar example would be (stolen from dlev, but this is how to make them truly comparable):
static void Function(ref string input)
{
input = "modified";
}
static void Function2(int[] val)
{
val = new int[1];
val[0] = 100;
}
Both functions change their parameter's pointer. Only because you used ref does input "remember" its changes, because when it changes the pointer, it is changing the pointer that was passed in and not just a copy of it.
val will still be an array of 10 ints outside of the function, and val[0] will still be 1 because the "val" within Function2 is a different pointer that originally points to the same location as Main's val, but it points somewhere else after the new array is created (the different pointer points to the new array, and the original pointer continues to point to the same location).
If I used ref with the int array, then it to would have changed. And it would have changed in size too.
A better example for newbies:
string a = "one";
string b = a;
string b = "two";
Console.WriteLine(a);
... will output "one".
Why? Because you are assigning a whole new string into pointer b.
The confusion is that ref type references are passed by value by default, to modify the reference itself (what the object points to) you have to pass the reference by reference - using ref.
In your case you are handling strings - assigning a string to a variable (or appending to, etc.) changes the reference, since strings are immutable there is no way to avoid this either, so you have to use ref.
You're right. Arrays and strings are reference types. But if to be honest and compare similar behavior you should write something like this:
static void Function2(int[] val) // It doesn't need 'ref' for a reference type
{
val = new[] { 1, 2, 3, 4 };
}
But in your example you perform a write operation in some element of a C# one-dimensional array via reference val.
I am passing a string variable to a method. I know strings are reference types but the value that I assign inside the method is lost.
public static void TestMethod(string myString)
{
myString = "world";
}
static void Main(string[] args)
{
string s = "hello";
Console.WriteLine(s); // output is "hello"
TestMethod(s);
Console.WriteLine(s); // output is also "hello" not "world" !?
}
Anyway this does not happen with an array for example. Can someone explain why might be the cause?
Because myString = "world" assigns a new string to the parameter, not updating the existing string. To update the original reference to the string you must pass the parameter with ref.
public static void TestMethod(ref string myString)
{
myString = "world";
}
static void Main(string[] args)
{
string s = "hello";
Console.WriteLine(s); // output is "hello"
TestMethod(ref s);
Console.WriteLine(s); // output is also "hello" not "world" !?
}
Yes, because without a ref (or out) you can't assign a new object to a parameter. Since you didn't pass it in through a ref, the variable outside the method still references the original string which didn't change. Consequently, strings are immutable, so you can't do anything to it without create a new string once it is instantiated.
The array can be altered (or the contents of the array can be altered), because the references inside the array are not immutable (you can say reassign my_object1 to equal "BLAH"). You can replace a value in the array and have it accessible outside of the array, because the reference to the array outside of the method hasn't changed.
Link to String in MSDN (talks about immutability)
For this to work you need to add the "ref" keyword to the parameter in the method signature.
Though your string is passed by reference, when you pass it to your method you have 2 references to the same string - the one in the Main() and the one in TestMethod(). When you assign a new value to the variable in the TestMethod() you are changing its reference, but not what Main()'s variable is referencing.
If you were able to just change the string from the TestMethod() instead of reassigning then you would see the effects in the Main(), but you can't with strings since they are immutable.
To play with this further you can try the following - change TestMethod() to receive a IList and add items to this list. You can see these new items in the variable you passed in from Main(). Now if you change TestMethod(IList listArg) to first reassign listArg to a new list (i.e. listArg = new List) and then add items, the list in Main() remains unchanged. This is the same idea.
Strings are immutable, which means that you can't change it's value, like you do with the items in an array. You can only replace a string object with a different string object.
The same happens if you try to replace an array object. This will put a new array in the parameter variable, but it will not change the variable that you used to call the method, and thus does not replace the array that you passed in:
public static void TestMethod(string[] myArray) {
myArray = new string[] { "world" };
}
Parameters are always passed by value unless you use the ref or out keywords. For a reference type that means that you pass a copy of the reference.
Using the ref keyword you pass the variable, so that you can change it in the method:
public static void TestMethod(ref string myString) {
myString = "world";
}
When reference variable can be passed by reference :
class Example
{
public string str="Demo";
public int[] intValues={1,3,4,5};
public static void StrPassing(string someStr)
{
string otherStr="Changed!";
someStr=otherStr;
}
public static void NumPassing(int[] a)
{
a[2] = 115;
}
}
static void Main(string[] args)
{
Example ex = new Example();
Example.StrPassing(ex.str);
Example.NumPassing(ex.intValues);
foreach (int i in ex.intValues)
{
Console.WriteLine(i);
}
Console.WriteLine(ex.str);
Console.ReadLine();
}
the value of intValues[2] is changed as 115 as the reference is being passed.But the value of the string "str" (demo) is not changed to "Changed!".What is the reason for it?.can i take it as Arrays are passed by reference and other reference types are passed by
value?
Whatever you pass to a method as arguments is passed by value which, for reference types, means that a reference is passed by value. So you can't change the object to another one but you can surely change its contents (because that doesn't change the actual reference, just some memory elsewhere).
As your example with the array demonstrates you take the array reference (but don't change it) and change a value in the array. This is just like taking some object and changing a property value. You can do this just fine from within a method too.
If you want to change a string, which is an immutable object in .NET, then you need to resort to ref parameters:
public static void StrPassing(ref string someStr)
{
string otherStr="Changed!";
someStr=otherStr;
}
And call it like this:
string foo = "foo";
StrPassing(ref foo);
Console.WriteLine(foo); // should print "Changed!"
The ref keyword ensures that your method gets the actual reference to the string and can change it, instead of just a copy of the reference. So then you can replace the object by an entirely new one.
To come back to your array: You'd have a hard time too, to change the passed array to an entirely different array:
public static void NumPassing(int[] a)
{
a = new int[15];
}
wouldn't work too because then you'd try exactly the same as changing a string to an entirely different string.
You need to distinguish between changing which object a variable refers to and changing *the content of the object".
In this code:
public static void StrPassing(string someStr)
{
string otherStr="Changed!";
someStr=otherStr;
}
... you are changing the value of someStr. You're not making any change to the string that someStr originally refers to. Indeed, you can't because strings are immutable. (If it were a StringBuilder, you could set the length to 0 and then append "Changed!")
Changing the value of someStr has no effect because the argument (ex.str) was passed by value. (The value in question is a reference, but that doesn't mean it's passed by reference.)
Now compare that with this code:
public static void NumPassing(int[] a)
{
a[2] = 115;
}
Here you're not changing the value of a - you're changing the contents of the array that a refers to.
In short, unless you use ref/out, arguments will be passed by value - but for reference types that value is just a reference.
I have an article on parameter passing which goes into all of this in a lot more detail.
What you'd need to do is change the signature for StrPassing to look like this:
public static void StrPassing(ref string someStr)
Strings are special in C#. They are immutable reference types which makes they exhibit similar behavior as value types.
Here's a good discussion.