Is it possible to set or override the default state for a structure?
As an example I have an
enum something{a,b,c,d,e};
and a struc
This may work...
public struct MyStruct
{
private bool _name;
public string myName
{
get { return (_name ? myName : "Default name"); }
set { _name = true; myName = value; }
}
private bool _num;
public int myNumber
{
get { return (_num ? myNumber : 42); }
set { _num = true; myNumber = value; }
}
private bool _bool;
public bool myBoolean
{
get { return (_bool ? myBoolean : true); }
set { _bool = true; myBoolean = value; }
}
private bool _type;
public MyRefType myType
{
get { return _type ? myType : new MyRefType(); }
set { _type = true; myType = value; }
}
}
Nevermind StackOverflowException
Creating a class object will cause all of the instance fields to come into existence before anything--even the class constructor--can access it, and allocating an array will cause all of its elements to exist before anything can access the array. Both of these actions will cause all of the memory allocated to those fields or elements to be zeroed out without regard for the data types to be stored therein.
When a class-type storage location comes into existence, it will initially hold a null reference. When a structure-type storage location comes into existence, all of its fields (and any fields of structures within it) will do so simultaneously. Unlike class object instances which can only come into existence by using a constructor, structure-type storage locations are brought into existence without using any of the structure's own code. Consequently, the structure's definition will have no say in what should happen when "instances" [i.e. struct-type storage locations] come into existence.
A struct is, fundamentally, a collection of fields bound together with duct tape. If a struct is supposed to behave like something else, it should typically make its fields private and pretend to be immutable [even though struct assignment actually mutates the destination struct by overwriting all its fields with the corresponding values from the source, and the struct definition gets no say in the matter]. If, however, a struct is supposed to encapsulate a fixed set of related but independent values (e.g. the coordinates of a point), which may independently accommodate any combination of values which are legal for their respective types, a struct should simply expose its fields publicly. Some people may whine about "mutable structs are evil", but the evils only apply when invoking self-mutating methods on a struct. Structs which expose their state as fields behave like collections of variables stuck together with duct tape. If what one needs is a collection of variables stuck together with duct tape, trying to make a struct pretend to be immutable will simply make it harder to program with.
You can't. Structs always have a default constructor that sets every member to its default value (null
for reference types, 0
for numeric types, false
for bools, etc.) This behavior cannot be changed.
Somewhat related: I've often wanted to use the new object initializer syntax with an immutable value type. However, given the nature of a typical immutable value type implementation, there is no way to utilize that syntax, since the properties are read-only.
I've come up with this approach; In my opinion this still satisfies the immutability of the value type, but allows the code that is responsible for instantiating the value type greater control over the initialization of the internal data.
struct ImmutableValueType
{
private int _ID;
private string _Name;
public int ID
{
get { return _ID; }
}
public string Name
{
get { return _Name; }
}
// Infuser struct defined within the ImmutableValueType struct so that it has access to private fields
public struct Infuser
{
private ImmutableValueType _Item;
// write-only properties provide the complement to the read-only properties of the immutable value type
public int ID
{
set { _Item._ID = value; }
}
public string Name
{
set { _Item._Name = value; }
}
public ImmutableValueType Produce()
{
return this._Item;
}
public void Reset(ImmutableValueType item)
{
this._Item = item;
}
public void Reset()
{
this._Item = new ImmutableValueType();
}
public static implicit operator ImmutableValueType(Infuser infuser)
{
return infuser.Produce();
}
}
}
class Program
{
static void Main(string[] args)
{
// use of object initializer syntax made possible by the Infuser type
var item = new ImmutableValueType.Infuser
{
ID = 123,
Name = "ABC",
}.Produce();
Console.WriteLine("ID={0}, Name={1}", item.ID, item.Name);
}
}
There is a workaround to make this happen by using custom Property getters. Observe:
public struct Foostruct
{
private int? _x;
private int? _y;
public int X
{
get { return _x ?? 20; } // replace 20 with desired default value
set { _x = value; }
}
public int Y
{
get { return _y ?? 10; } // replace 10 with desired default value
set { _y = value; }
}
}
This will only work for value types (which can be wrapped with nullable) but you could potentially do something similar for reference types by wrapping them in a generic class like below:
public class Wrapper<TValue>
{
public TValue Value { get; set; }
}
public struct Foostruct
{
private Wrapper<Tick> _tick;
public Tick Tick
{
get { return _tick == null ? new Tick(20) : _tick.Value; }
set { _tick = new Wrapper<Tick> { Value = value }; }
}
}
My solution. It works as well.
public struct DisplayOptions
{
public bool isUpon;
public bool screenFade;
public static DisplayOptions Build()
{
// Return default value
return new DisplayOptions(true, true);
}
DisplayOptions(bool isUpon, bool screenFade)
{
this.isUpon = isUpon;
this.screenFade = screenFade;
}
public DisplayOptions SetUpon(bool upon)
{
this.isUpon = upon;
return this;
}
public DisplayOptions SetScreenFade(bool screenFade)
{
this.screenFade = screenFade;
return this;
}
}
Use default value
// Use default
UIMaster.Instance.StartScreen("Screen 2", DisplayOptions.Build());
// Use custome
UIMaster.Instance.StartScreen("Screen 2", DisplayOptions.Build().SetScreenFade(false));
UIMaster.Instance.StartScreen("Screen 2", DisplayOptions.Build().SetUpon(false));