Single-value object initializer

Question

This might be lame, but here:

public interface Interface<T>
{
    T Value { get; }
}

public class InterfaceProxy<T> : Interface<T>
{
    public T Value { get; set; }
}

public class ImplementedInterface: InterfaceProxy<Double> {}

Now I want to create an instance of the ImplementedInterface and initialize its members.

Can this be done like this (using an object initializer) or can this only be achieved by using a constructor with Double argument?

var x = new ImplementedInterface { 30.0 };

Definitely not lame, nice question. +1

Gabe
– Gabe

2010-12-07 16:27:26 +00:00
Commented Dec 7, 2010 at 16:27 — Gabe
– Gabe, Commented Dec 7, 2010 at 16:27

Saeed Amiri · Accepted Answer · 2010-12-07 16:24:37Z

7

Can be done by:

var x = new ImplementedInteface { Value = 30.0 };

answered Dec 7, 2010 at 16:24

Saeed Amiri

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6 Comments

Marcel Gheorghita Over a year ago

You missed the round brackets after class name.

Saeed Amiri Over a year ago

@Marcel Gheorghita, you can skip this bracket. no need.

Gabe Over a year ago

@Marcel - those are optional.

Brian Ball Over a year ago

If using a parameterless constructor, it's been optional since object initialization was introduced.

Saeed Amiri Over a year ago

@Marcel Gheorghita, no marcel, as Brian said, It comes with object initialization, if you have resharper, offers to remove this brackets.

|

Marcel Gheorghita · Accepted Answer · 2010-12-07 16:25:20Z

2

var x = new ImplementedInterface() { Value = 30.0 };

answered Dec 7, 2010 at 16:25

Marcel Gheorghita

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Comments

BFree · Accepted Answer · 2010-12-07 16:28:01Z

2

The only way to achieve what you're after is if your class implements IEnumerable<T> and has an Add method:

public class MyClass : IEnumerable<double>
{
   public void Add(double x){}
}

Then you can do:

MyClass mc = new MyClass { 20.0 };

Obviously that's not what you want, because that doesn't set your Value and it allows you to add multiple values:

MyClass mc = new MyClass { 20.0, 30.0 , 40.0 };

Just go with the standard object initializes like others have pointed out:

var x = new ImplementedInterface() { Value = 30.0 };

answered Dec 7, 2010 at 16:28

BFree

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Comments

Brian Ball · Accepted Answer · 2013-03-04 15:02:12Z

2

You should be able to do:

var x = new ImplementedInterface {Value = 30.0};

edited Mar 4, 2013 at 15:02

user142162

answered Dec 7, 2010 at 16:24

Brian Ball

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Comments

user197015 · Accepted Answer · 2010-12-07 16:28:14Z

1

var instance = new ImplementedInterface { Value = 30.0 }; will work. However, this isn't really the same set of operations as C++ initializer lists -- this is an object initializer. It initializes the new instance via the default constructor and then invokes the property setters for each property.

In other words, the object is constructed before the property setters run. If you want the values for the properties set before construction of ImplementedInterface completes, you'd have to write a constructor, as you noted. This distinction in behavior usually doesn't matter, but it's good to be aware of.

answered Dec 7, 2010 at 16:28

user197015

Comments

Kadir Sümerkent · Accepted Answer · 2010-12-07 16:28:36Z

1

I am not sure if you have a special reason to use the interfaces that way but the following code might work for you.

public class ImplementedInterface2 : List<double> { }

 public class test
    {
        public void x()
        {
            var x = new ImplementedInterface2() { 30.0 };
        }
    }

answered Dec 7, 2010 at 16:28

Kadir Sümerkent

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Comments

Ricky Smith · Accepted Answer · 2010-12-07 16:25:32Z

0

var x = new ImplementedInterface { Value = 30.0 };

answered Dec 7, 2010 at 16:25

Ricky Smith

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Comments

riwalk · Accepted Answer · 2010-12-07 16:27:08Z

0

You can definitely use an initialization list, but you have to specify what 30.0 is (this is true for any initialization list, not just the code you have):

var x = new ImplementedInteface { Value=30.0 };

answered Dec 7, 2010 at 16:27

riwalk

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