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1712

answers:

4

I know that "string" in C# is a reference type. This is on MSDN. However, this code doesn't work as it should then:

class Test
{
    public static void Main()
    {
        string test = "before passing";
        Console.WriteLine(test);
        TestI(test);
        Console.WriteLine(test);
    }

    public static void TestI(string test)
    {
        test = "after passing";
    }
}

The output should be "before passing" "after passing" since I'm passing the string as a parameter and it being a reference type, the second output statement should recognize that the text changed in the TestI method. However, I get "before passing" "before passing" making it seem that it is passed by value not by ref. I understand that strings are immutable, but I don't see how that would explain what is going on here. What am I missing? Thanks.

+18  A: 

The reference to the string is passed by value. There's a big difference between passing a reference by value and passing an object by reference. It's unfortunate that the word "reference" is used in both cases.

If you do pass the string reference by reference, it will work as you expect:

using System;

class Test
{
    public static void Main()
    {
        string test = "before passing";
        Console.WriteLine(test);
        TestI(ref test);
        Console.WriteLine(test);
    }

    public static void TestI(ref string test)
    {
        test = "after passing";
    }
}

Now you need to distinguish between making changes to the object which a reference refers to, and making a change to a variable (such as a parameter) to let it refer to a different object. We can't make changes to a string because strings are immutable, but we can demonstrate it with a StringBuilder instead:

using System;
using System.Text;

class Test
{
    public static void Main()
    {
        StringBuilder test = new StringBuilder();
        Console.WriteLine(test);
        TestI(test);
        Console.WriteLine(test);
    }

    public static void TestI(StringBuilder test)
    {
        // Note that we're not changing the value
        // of the "test" parameter - we're changing
        // the data in the object it's referring to
        test.Append("changing");
    }
}

See my article on parameter passing for more details.

Jon Skeet
agree, just want to make clear that using the ref modifier also works for non-reference types i.e. both are pretty separate concepts.
eglasius
+1 nice article
Chalkey
+2  A: 

Try:


public static void TestI(ref string test)
    {
        test = "after passing";
    }
Marius Kjeldahl
+4  A: 

Actually it would have been the same for any object for that matter i.e. being a reference type and passing by reference are 2 different things in c#.

This would work, but that applies regardless of the type:

public static void TestI(ref string test)

Also about string being a reference type, its also a special one. Its designed to be immutable, so all of its methods won't modify the instance (they return a new one). It also has some extra things in it for performance.

eglasius
A: 

If we have to answer the question: String is a reference type and it behaves as a reference. We pass a parameter that holds a reference to, not the actual string. The problem is in the function:

public static void TestI(string test)
{
    test = "after passing";
}

The parameter test holds a reference to the string but it is a copy. We have two variables pointing to the string. And because any operations with strings actually create a new object, we make our local copy to point to the new string. But the original test variable is not changed.

The suggested solutions to put ref in the function declaration and in the invocation work because we will not pass the value of the test variable but will pass just a reference to it. Thus any changes inside the function will reflect the original variable.

I want to repeat at the end: String is a reference type but since its immutable the line test = "after passing"; actually creates a new object and out copy of the variable test is changed to point to the new string.

Martin Dimitrov