12

I have been trying to figure out if there are any differences between these constructors. Assuming there is a Foo() constructor that takes no arguments, are all these constructors going to have the same result?

Example 1

public Foo()
    : this()
{
     blah;
     blah;
     blah;
}

Example 2

public Foo()
{
     this();
     blah;
     blah;
     blah;
}

Example 3

public Foo()
{
     this = new Foo();
     blah;
     blah;
     blah;
}
1
  • 2
    +1 taught me something new with Example#3 n the fact that J S is always lurking Commented Jun 16, 2009 at 14:24

3 Answers 3

34
  • Example 1 is valid (assuming there is a parameterless constructor), and calls the parameterless constructor as part of initialization. See my article on constructor chaining for more details. EDIT: Note that since the OP's edit, it's infinitely recursive.
  • Example 2 is never valid
  • Example 3 is only valid when Foo is a struct, and doesn't do anything useful.

I would steer clear of assigning to this in structs. As you can see from the other answers, the very possibility of it is fairly rarely known (I only know because of some weird situation where it turned up in the spec). Where you've got it, it doesn't do any good - and in other places it's likely to be mutating the struct, which is not a good idea. Structs should always be immutable :)

EDIT: Just to make people go "meep!" a little - assigning to this isn't quite the same as just chaining to another constructor, as you can do it in methods too:

using System;

public struct Foo
{
    // Readonly, so must be immutable, right?
    public readonly string x;

    public Foo(string x)
    {
        this.x = x;
    }

    public void EvilEvilEvil()
    {
        this = new Foo();
    }
}

public class Test
{
    static void Main()
    {
        Foo foo = new Foo("Test");
        Console.WriteLine(foo.x); // Prints "Test"
        foo.EvilEvilEvil();
        Console.WriteLine(foo.x); // Prints nothing
    }
}
Sign up to request clarification or add additional context in comments.

13 Comments

I wasn't aware 3 was legal, but you're right, it is. I always accomplished the same thing by chaining the parameterless constructor.
I had always thought that : was used for inheritance - is there any inheritance happening in Example 1, or is this just an overloaded usage of the colon?
The colon just means something different in this context. Constructor chaining on MSDN: msdn.microsoft.com/en-us/library/ms173115.aspx
That example is quite remarkable. I'm very surprised that you can do such a thing.
@mquander: Yes, it's thoroughly evil, isn't it? Ick.
|
11

Examples 2 and 3 are not legal C#.

EDIT: Jon points out accurately that 3 is legal when Foo is a struct. Go check out his answer!

1 Comment

#3 is used as a copy constructor struct Foo { int a; Foo(Foo other) { this=other; } }
4

No they will not because only the first constructor is actually legal. The other two are illegal for various reasons.

EDIT Interesting, 3 is indeed legal when Foo is a struct. But even in that case, it is a redundant assignment.

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.