How to pass in the "modifier" string returned by the function?

Question

class test
{
  public String get() {return s;}
  private String s="World";
}

class modifier
{
   public static void modify(ref String v)
   {
     v+="_test";
   }
}

String s1="Earth";


modifier.modify(ref s1); <-------- OK

test c=new test();
modifier.modify(ref c.get()); <------- Error

How to pass in the "modifier" string returned by the function? Assignment through another String object is unacceptable. So how will the copy is created.

Remember: strings are _immutable_ types. You **can't** change them. — Soner Gönül, Jan 28 '13 at 14:17
public String get() {return s;} private String s="World"; not sure but try changing the order of this two put the private first — Christopher Cabezudo Rodriguez, Jan 28 '13 at 14:20
"ref" allows me to pass a string as a parameter and change it. is there no such a mechanism to obtain a String as the result of the function? — Mixer, Jan 28 '13 at 14:21
public static string modify(String v) { return v + "_test"; } — Christopher Cabezudo Rodriguez, Jan 28 '13 at 14:23
Just return the new string. Note that when you pass a string by ref, you are returning a NEW string via the ref (the old string disappears). — Matthew Watson, Jan 28 '13 at 14:24
String internaly will always create another object each time that u change his content. if you dont want that behavior try using stringbuilder — Christopher Cabezudo Rodriguez, Jan 28 '13 at 14:24

score 2 · Answer 1 · answered Jan 28 '13 at 14:30

You're trying to write C++ code in C#. You'll need to approach this the C# way. Add a property for test.s, instead of a method called "get()". We'll call this new property "MyString" in the code below:

class test
{
    public String MyString
    {
        get
        {
            return s;
        }

        set
        {
            s = value;
        }
    }

    private String s = "World"; 
}

class modifier
{
    public static string modify(String v)
    {
        return v + "_test";
    }
}

test c = new test();
c.MyString = modifier.modify(c.MyString);

George · Accepted Answer · 2013-01-28T19:01:18.183

If you dont like Matthew Watson's answer and you want to stay to your approach, the way to go is by using StringBuilder.That is a String that can change its value without creating a new one!

What i mean:
Normal String's value cant be changed, what is happening is that a new String is being created....and the pointer(that points to the String you want to change its value) points from now on to this new String. All other pointers....that "were pointing" to the old String .....are still pointing...to the old String!(the String's value didnt change!)
I am not sure if thats clear enough but you have to understand this if you want to play with Strings.This is exactly why s1 cant change its value.

The workaround is with StringBuilder:

    class test
{
    public StringBuilder get() { return  s; }
    private StringBuilder s = new StringBuilder("World");
}

class modifier
{
    public static void modify(StringBuilder v)
    {
        v.Append("_test");
    }
}

And some test code : (of course all this comes with processing cost...but i dont think that will be an issue for now)

        StringBuilder s1 = new StringBuilder("Earth");

        System.Diagnostics.Debug.WriteLine("earth is {0}", s1);
        modifier.modify(s1); //<-------- OK
        System.Diagnostics.Debug.WriteLine("earth is {0}",s1);

        test c=new test();
        StringBuilder aa=c.get();
        System.Diagnostics.Debug.WriteLine("earth is {0}", aa);
        modifier.modify(aa); //<------- Error(not anymore)
        System.Diagnostics.Debug.WriteLine("earth is {0}", c.get());

Before you use the code try to understand how String and StringBuilder works

Hope it helps you clear things up. Its very important to understand why your code is not working before going on. — George, Jan 28 '13 at 15:06
In this scenario, the parameter to the `modify` method probably shouldn't be `ref`. After all, the parameter is not reassigned inside the method body, and it's a reference type. — Jeppe Stig Nielsen, Jan 28 '13 at 15:20

score 0 · Answer 3 · answered Jan 28 '13 at 14:17

0

You have to create a temporary variable:

string tmp = c.get();
modifier.modify(ref tmp);

Because you are passing your parameter by reference.

answered Jan 28 '13 at 14:17

Cédric Bignon

12,892
3
39
51

I can't create a copy. I need to change the String of the test class. – Mixer Jan 28 '13 at 14:19
2

Then, add an _set()_ method to the _test_ class. – Cédric Bignon Jan 28 '13 at 14:20
@Mixer You can't change strings. Strings are immutable types. – JustAnotherUserYouMayKnow Jan 28 '13 at 14:22
Result of a function does not occupy a particular space in memory unless it is assigned to a variable. It can't be done. – Darek Jan 28 '13 at 14:27
2

@JustAnotherUserYouMayKnow When people say "change the string", obviously what they really mean is "change the _reference_ of the field (or variable) to refer another string instance". – Jeppe Stig Nielsen Jan 28 '13 at 14:33

score 0 · Answer 4 · answered Jan 28 '13 at 14:29

The class test is designed such that its field s can't be reassigned (changed to point to a new object) from outside the class. That's because s is private, and the get() method only returns s, hence can't reassign it.

Either change the class test to allow the outside world to reassign s somehow, or use reflection to access a private field from the outside.

score 0 · Answer 5 · answered Jan 28 '13 at 14:49

You can use a property with a get/set-eccessor and then pass to the modify() a test object:

class test
{
    public String myString 
    { 
         get { return s; } 
         set { s = value; } 
    }

    private String s="World";
}

class modifier
{
    public static void modify(test myTest)
    {
        myTest.myString += "_test";
    } 
}

test c = new test();
modifier.modify(c);

Console.WriteLine(c.myString); //World_test

How to pass in the "modifier" string returned by the function?

5 Answers5