Consider the following code:
public struct Color {
public int R;
public int G;
public int B;
}
public class App
{
static void Main()
{
Color c;
c.B = 0xFF;
int b = c.B;
}
}
csc compiles the code happily. I always thought all fields of a struct have to be assigned to before one can access data members of the struct. Is this a specialty of csc.exe?
I think NullReferenceExceptions are not the right solution, since we are talking about structs here.
From MSDN:
When you create a struct object using the new operator, it gets
created and the appropriate constructor is called. Unlike classes,
structs can be instantiated without using the new operator. In such a
case, there is no constructor call, which makes the allocation more
efficient. However, the fields will remain unassigned and the object
cannot be used until all of the fields are initialized.
From MSDN:
Compiler Error CS0170: Use of possibly unassigned field 'field'. A
field in a structure was used without first being initialized. To
solve this problem, first determine which field was uninitialized and
then initialize it before you try to access it.
From MSDN:
Compiler Error CS0165: Use of unassigned local variable 'name'. The C# compiler does not allow the use of uninitialized variables. If
the compiler detects the use of a variable that might not have been
initialized, it generates compiler error CS0165.
This is wrong:
I always thought all fields of a struct have to be assigned to before one can access data members of the struct
The correct one is:
All fields of a struct have to be assigned to before one can access the struct.
Color c;
c.B = 0xFF;
int b = c.B; // Okay. You have assigned B
int r = c.R; // Error CS0170! Use of possibly unassigned field
Color cc = c; // Error CS0165! Use of unassigned local variable. The object cannot be used until all of the fields are initialized
Refer this link
msdn ink
If you use Color c; fields are not initialized, but if you do Color c = new Color(); all fields will be initialized.
if you run the below code. it will failed to compile.
Color c;
int b = c.B;
But this will be compiled.
Color c = new Color();
// c.B = 0xFF;
int b = c.B;
Related
The following C# struct is used to represent a union of color components and the 32bit color value itself. The problem is that the compiler gives the error:
Error CS0171 Field 'Color.ARGB' must be fully assigned before control is returned to the caller
Is it possible to get rid of this error without initialize the data twice? Is this expected behavior of C#? If I init twice, will the JIT detect the dual init and only do the second one?
[StructLayout(LayoutKind.Explicit)]
public struct Color
{
public Color(byte r, byte g, byte b, byte a = 0xff)
{
ARGB = 0; // The init I shouldn't have to do
A = a;
R = r;
G = g;
B = b;
}
[FieldOffset(0)]
public byte B;
[FieldOffset(1)]
public byte G;
[FieldOffset(2)]
public byte R;
[FieldOffset(3)]
public byte A;
[FieldOffset(0)]
public uint ARGB;
public static readonly Color Red = new Color(255, 0, 0);
public static readonly Color Green = new Color(0, 255, 0);
public static readonly Color Blue = new Color(0, 0, 255);
}
Is it possible to get rid of this error without initialize the data twice?
Yes and no.
The supposition here is that the members are not already "initialized twice". When you get the new struct from the memory allocator -- either from the heap or the stack -- it will be automatically zeroed out.
As Naidu's answer notes, calling the default constructor indicates to the compiler "the runtime must zero this thing out if it is not already; I wish to assert that I am fine with any portion of the object not written to by the constructor being left in its default state ".
In practice, typically the jitter has already initialized to zero, so typically there is no extra initialization done. However, the behaviour that memory allocators automatically initialize state to zero is runtime-implementation-dependent. Similarly it is an implementation-dependent behaviour whether or not the jitter can optimize away the zero-out behaviour if it knows that every field is initialized.
There are subtleties here. Suppose for example the memory is not zeroed out because the jitter has deduced that your constructor writes every field. Now suppose a thread abort exception is thrown halfway through the constructor. Is it possible for another thread to observe the not-zeroed-out, not-written-by-you state of the object? What hellish behaviour might that wreak, if in fact it is possible? Give that some thought.
Is this expected behavior of C#?
Yes.
The compiler has no idea whatsoever that you're creating a type-unsafe union. It doesn't know the meanings of those attributes.
If I init twice, will the JIT detect the dual init and only do the second one?
There are many different jitters on many different platforms. If you want an answer to your question, try it on all of them with all possible configurations and see what happens.
Regardless, you are likely worrying about nothing important. Writing zeros into memory is pretty fast. Doing an unnecessary zero write is probably not the bottleneck in your program.
Look into below Microsoft link:
https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/compiler-messages/cs0843
It says,
To assign a value to an automatically-implemented property from a
constructor, you must first invoke the default constructor to create
the object.
Doing below change will resolve your issue. call default constructor.
public Color(byte r, byte g, byte b, byte a = 0xff):this()
{
A = a;
R = r;
G = g;
B = b;
}
Use System.Runtime.CompilerServices.Unsafe.SkipInit, available since .NET 5. It is supposed to actually skip zero-initialization for bytes you're going to initialize anyway:
public Color(byte r, byte g, byte b, byte a = 0xff)
{
Unsafe.SkipInit(out ARGB);
A = a;
R = r;
G = g;
B = b;
}
For static fields, circular dependencies result in the default value.
static int a = b; // 0
static int b = a; // 0
It doesn't apply to instance fields though, as they cannot reference each other.
int c = d; // CS0236 A field initializer cannot reference the non-static field, method, or property
int d = c; // CS0236
Why is this allowed for static fields?
It is not circular dependency but the difference in how static and instance fields are initialized.
A variable initializer for an instance field cannot reference the
instance being created. Thus it is a compile-time error to reference
this in a variable initializer, because it is a compile-time error for
a variable initializer to reference any instance member through a
simple-name
In short, you can't reference instance variable before it is created.
Update: I saw this question with +4 votes. Never thought it's such an easy question. I misunderstood the question, but I think my answer is still helpful so I'm not going to delete it.
It's not a circular dependencies problem. Why the first example is allowed? Just consider this:
In a method, this is ok:
int a;
int b;
a = b;
b = a;
but this is not allowed:
int c;
c = d;
int d;
d = c;
And the order of declarations of fields/methods in a class is not important (but definitions have order)
This question already has answers here:
Why can I not modify the result of an unboxing conversion?
(3 answers)
Closed 7 years ago.
Let's look at the following code.
struct SPoint
{
public int x;
public int y;
public SPoint(int x, int y)
{
this.x = x;
this.y = y;
}
}
class Test
{
public static void Main()
{
SPoint s = new SPoint(3, 4);
object o = s;
((SPoint) o).x = 5;
}
}
Why isn't the last assignment possible? What is the reason for such behaviour?
Since s is a struct (a.k.a: a value type), (SPoint)o is a copy of the data:
From the C# language spec (§1.3, "Types and Variables"):
When a value of a value type is converted to type object, an object instance, also called a “box,” is allocated to hold the value, and the value is copied into that box. Conversely, when an object reference is cast to a value type, a check is made that the referenced object is a box of the correct value type, and, if the check succeeds, the value in the box is copied out.
The language protects you from changing the data of the cloned value type without putting it in a variable first, since you might think that you are changing the original s.x value while you are changing it's temporary (not - variabled) clone, unlike unsafe languages like C++/CLI where this kind of assignments might be allowed.
If you want, you are able to explicitly create a new variable and do your manipulations in it:
SPoint cloneOfS = ((SPoint)o);
cloneOfS.x = 5;
See MSDN:
The result of an unboxing conversion is a temporary variable. The compiler prevents you from modifying such variables because any modification would go away when the temporary variable goes away. To fix this, declare a new value-type variable to store the intermediate expression, and assign the result of the unboxing conversion to that variable.
In Visual Studio 2012 i have this simple example here:
and when i debug the code and move my cursor over i to get it's current value, i expect something like
Use of unassigned local variable
but there is a 0 which was not set - why is there a 0?
When you declare any local/block variable, they didn’t get the default values. They must assigned some value before accessing it other wise compiler will throw an error. If the variable has a global scope then default value can be assigned and accessed. if the variable is of reference type then the default value will be null. this link contains default values for the primitive datatype:
The compiler will not permit this(since it is local/block variable):
public static void samplemethod()
{
int a;
int b = a;
}
where as the following code works fine since the variable have global scope:
public int i;
public void samplemethod()
{
int a;
int b = i;
}
That's because int is a Value Type not a Reference Type.
MSDN :
Variables that are based on value types directly contain values.
Assigning one value type variable to another copies the contained
value. This differs from the assignment of reference type variables,
which copies a reference to the object but not the object itself.
Have a look at Value Types and Reference Types.
I hope it turns out to be helpful.
I started my journey with C# but I realised that I have some problems with some basic information about memory when it comes to declaration of variables. See if I am correct.
int x; // I declared variable of type int, which name is x. Compiler will provide memory for it but we dont have known value of it.
x=10; // Now memory location is still the same but value now kept there is 10;
public struct Point {
public int x, y;
}
Now I define a struct named Point. Beacuse struct is a value type, it again has reserved memory for it on the computer. Howewer x and y have no value.
Now Point p1 = new Point(); // what is happening here? Struct is not a reference type. So is this just initialization of Point variable with the default constructor without assigning values to x and y?
Second short question. When I write a code like:
int x = 10;
Can I say that I created instance of class integer which value is 10 and name x;
I would be grateful for help.
// what is happening here? Struct is not a reference type. So is this just initialization of Point variable with the default constructor without assigning values to x and y?
No; there are 4 possible scenarios here:
a class: the memory space is wiped to all 0s, then any custom constructor is invoked, which may also involve field initializers
a struct called without a custom constructor: the memory space is wiped to all 0s
a struct called with a custom constructor: the custom constructor is required to assign all the fields
a struct variable used without ever calling a constructor: this is actually a thing, but the calling code must write to all the fields before they can do anything else with it; since most structs do not expose their fields, this rarely works
Second short question. When i write a code like:
int x = 10;
Can i say that i created instance of class integer which value is 10 and name x; I would be grateful for help.
Not really, because in C# terms, int is not a class (it might be in IL terms). Simply: you have declared a local variable of type int with name x and assigned it the value 10, if this is in a method. If this is a class field, then: you have declared a private instance field of type int named x with a field-initializer giving it the value of 10.
Incidentally, you should avoid public fields in general, and mutable fields on structs. You might prefer:
public struct Point {
private readonly int x, y;
public int X { get { return x; } }
public int Y { get { return y; } }
public Point(int x, int y) { this.x = x; this.y = y'; }
}
This will avoid a huge range of problems.
In C# the default struct constructor sets the struct memory to 0, effectively setting all variables to their default values.
In case of ints, it will be 0. For reference types, it will result in null.
(in other words, for any type T it will be default(T)).
Note that when you write a custom constructor in a struct, you must initialize all member fields.
When you write
int x;
this is similar to
Point p1 = new Point(); (considering Point structure is already defined)
in both the cases all integer variables will have default value of 0 and not null, which is is basically what is used in C# to denote 'nothing' and can be assigned only to reference types.
As well, in c# everything is a class, so when you write
int x = 10;
you are creating an instance of class Int32, though the run time will handle this as value type instead of ref type, as special case.
Same is true for other basic types like, Long, DateTime and few others