C# operator overload for `+=`?

I am trying to do operator overloads for +=, but I can't. I can only make an operator overload for +.

How come?

Edit

The reason this is not working is that I have a Vector class (with an X and Y field). Consider the following example.

vector1 += vector2;

If my operator overload is set to:

public static Vector operator +(Vector left, Vector right)
{
return new Vector(right.x + left.x, right.y + left.y);
}

Then the result won't be added to vector1, but instead, vector1 will become a brand new Vector by reference as well.

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最佳答案

Overloadable Operators, from MSDN:

Assignment operators cannot be overloaded, but +=, for example, is evaluated using +, which can be overloaded.

Even more, none of assignment operators can be overloaded. I think this is because there will be an effect for the Garbage collection and memory management, which is a potential security hole in CLR strong typed world.

Nevertheless, let's see what exactly an operator is. According to the famous Jeffrey Richter's book, each programming language has its own operators list, which are compiled in a special method calls, and CLR itself doesn't know anything about operators. So let's see what exactly stays behind the + and += operators.

See this simple code:

Decimal d = 10M;
d = d + 10M;
Console.WriteLine(d);

Let view the IL-code for this instructions:

  IL_0000:  nop
IL_0001:  ldc.i4.s   10
IL_0003:  newobj     instance void [mscorlib]System.Decimal::.ctor(int32)
IL_0008:  stloc.0
IL_0009:  ldloc.0
IL_000a:  ldc.i4.s   10
IL_000c:  newobj     instance void [mscorlib]System.Decimal::.ctor(int32)
IL_0011:  call       valuetype [mscorlib]System.Decimal [mscorlib]System.Decimal::op_Addition(valuetype [mscorlib]System.Decimal,
valuetype [mscorlib]System.Decimal)
IL_0016:  stloc.0

Now lets see this code:

Decimal d1 = 10M;
d1 += 10M;
Console.WriteLine(d1);

And IL-code for this:

  IL_0000:  nop
IL_0001:  ldc.i4.s   10
IL_0003:  newobj     instance void [mscorlib]System.Decimal::.ctor(int32)
IL_0008:  stloc.0
IL_0009:  ldloc.0
IL_000a:  ldc.i4.s   10
IL_000c:  newobj     instance void [mscorlib]System.Decimal::.ctor(int32)
IL_0011:  call       valuetype [mscorlib]System.Decimal [mscorlib]System.Decimal::op_Addition(valuetype [mscorlib]System.Decimal,
valuetype [mscorlib]System.Decimal)
IL_0016:  stloc.0

They are equal! So the += operator is just syntactic sugar for your program in C#, and you can simply overload + operator.

For example:

class Foo
{
private int c1;


public Foo(int c11)
{
c1 = c11;
}


public static Foo operator +(Foo c1, Foo x)
{
return new Foo(c1.c1 + x.c1);
}
}


static void Main(string[] args)
{
Foo d1 =  new Foo (10);
Foo d2 = new Foo(11);
d2 += d1;
}

This code will be compiled and successfully run as:

  IL_0000:  nop
IL_0001:  ldc.i4.s   10
IL_0003:  newobj     instance void ConsoleApplication2.Program/Foo::.ctor(int32)
IL_0008:  stloc.0
IL_0009:  ldc.i4.s   11
IL_000b:  newobj     instance void ConsoleApplication2.Program/Foo::.ctor(int32)
IL_0010:  stloc.1
IL_0011:  ldloc.1
IL_0012:  ldloc.0
IL_0013:  call       class ConsoleApplication2.Program/Foo ConsoleApplication2.Program/Foo::op_Addition(class ConsoleApplication2.Program/Foo,
class ConsoleApplication2.Program/Foo)
IL_0018:  stloc.1

Update:

According to your Update - as the @EricLippert says, you really should have the vectors as an immutable object. Result of adding of the two vectors is a new vector, not the first one with different sizes.

If, for some reason you need to change first vector, you can use this overload (but as for me, this is very strange behaviour):

public static Vector operator +(Vector left, Vector right)
{
left.x += right.x;
left.y += right.y;
return left;
}
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I think you'll find this link informative: Overloadable Operators

Assignment operators cannot be overloaded, but +=, for example, is evaluated using +, which can be overloaded.

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This is because of this operator can't be overloaded:

Assignment operators cannot be overloaded, but +=, for example, is evaluated using +, which can be overloaded.

MSDN

Just overload + operator, because of

x += y equal to x = x + y

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This is because of the same reason that the assignment operator cannot be overloaded. You cannot write code that would perform the assignment correctly.

class Foo
{
// Won't compile.
public static Foo operator= (Foo c1, int x)
{
// duh... what do I do here?  I can't change the reference of c1.
}
}

Assignment operators cannot be overloaded, but +=, for example, is evaluated using +, which can be overloaded.

From MSDN.

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Operator overload for + is used in += operator, A += B equals to A = operator+(A, B).

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If you overload + operator like this: 

class Foo
{
public static Foo operator + (Foo c1, int x)
{
// implementation
}
}

you can do

 Foo foo = new Foo();
foo += 10;

or 

 foo = foo + 10;

This will compile and run equally.

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You can't overload += because it isn't really a unique operator, it's just syntactic sugar. x += y is just a shorthand way of writing x = x + y. Because += is defined in terms of the + and = operators, allowing you to override it separately could create problems, in cases where x += y and x = x + y didn't behave in the exact same way.

At a lower level, it's very likely that the C# compiler compiles both expressions down to the same bytecode, meaning that it's very likely the the runtime can't treat them differently during program execution.

I can understand that you might want to treat it like a separate operation: in an statement like x += 10 you know that you can mutate the x object in place and perhaps save some time/memory, rather than creating a new object x + 10 before assigning it over the old reference.

But consider this code:

a = ...
b = a;
a += 10;

Should a == b at the end? For most types, no, a is 10 more than b. But if you could overload the += operator to mutate in place, then yes. Now consider that a and b could get passed around to distant parts of the program. Your possible optimization could create confusing bugs if your object starts to change where code doesn't expect it to.

In other words, if performance is that important, it's not too hard to replace x += 10 with a method call like x.increaseBy(10), and it's a lot clearer for everyone involved.

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I had the exact same question and i can not possibly answer it better then this person has

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There is always the same answer to this problem: Why do you need the +=, if you get it for free if you overload the +. But what happens if I have a class like this.

using System;
using System.IO;


public class Class1
{
public class MappableObject
{
FileStream stream;


public  int Blocks;
public int BlockSize;


public MappableObject(string FileName, int Blocks_in, int BlockSize_in)
{
Blocks = Blocks_in;
BlockSize = BlockSize_in;


// Just create the file here and set the size
stream = new FileStream(FileName); // Here we need more params of course to create a file.
stream.SetLength(sizeof(float) * Blocks * BlockSize);
}


public float[] GetBlock(int BlockNo)
{
long BlockPos = BlockNo * BlockSize;


stream.Position = BlockPos;


using (BinaryReader reader = new BinaryReader(stream))
{
float[] resData = new float[BlockSize];
for (int i = 0; i < BlockSize; i++)
{
// This line is stupid enough for accessing files a lot and the data is large
// Maybe someone has an idea to make this faster? I tried a lot and this is the simplest solution
// for illustration.
resData[i] = reader.ReadSingle();
}
}


retuen resData;
}


public void SetBlock(int BlockNo, float[] data)
{
long BlockPos = BlockNo * BlockSize;


stream.Position = BlockPos;


using (BinaryWriter reader = new BinaryWriter(stream))
{
for (int i = 0; i < BlockSize; i++)
{
// Also this line is stupid enough for accessing files a lot and the data is large
reader.Write(data[i];
}
}


retuen resData;
}


// For adding two MappableObjects
public static MappableObject operator +(MappableObject A, Mappableobject B)
{
// Of course we have to make sure that all dimensions are correct.


MappableObject result = new MappableObject(Path.GetTempFileName(), A.Blocks, A.BlockSize);


for (int i = 0; i < Blocks; i++)
{
float[] dataA = A.GetBlock(i);
float[] dataB = B.GetBlock(i);


float[] C = new float[dataA.Length];


for (int j = 0; j < BlockSize; j++)
{
C[j] = A[j] + B[j];
}


result.SetBlock(i, C);
}
}


// For adding a single float to the whole data.
public static MappableObject operator +(MappableObject A, float B)
{
// Of course we have to make sure that all dimensions are correct.


MappableObject result = new MappableObject(Path.GetTempFileName(), A.Blocks, A.BlockSize);


for (int i = 0; i < Blocks; i++)
{
float[] dataA = A.GetBlock(i);


float[] C = new float[dataA.Length];


for (int j = 0; j < BlockSize; j++)
{
C[j] = A[j] + B;
}


result.SetBlock(i, C);
}
}


// Of course this doesn't work, but maybe you can see the effect here.
// when the += is automimplemented from the definition above I have to create another large
// object which causes a loss of memory and also takes more time because of the operation -> altgough its
// simple in the example, but in reality it's much more complex.
public static MappableObject operator +=(MappableObject A, float B)
{
// Of course we have to make sure that all dimensions are correct.


MappableObject result = new MappableObject(Path.GetTempFileName(), A.Blocks, A.BlockSize);


for (int i = 0; i < Blocks; i++)
{
float[] dataA = A.GetBlock(i);


for (int j = 0; j < BlockSize; j++)
{
A[j]+= + B;
}


result.SetBlock(i, A);
}
}
}
}

Do you still say that it is good that the += is "auto-implemented". If you try to do high-performance computing in C# you need to have such features to reduce processing time and memory consumption, if someone has a good solution it is highly appreciated, but don't tell me I have to do this with static methods, this is only a workaround and I see no reason why C# does the += implemention if it is not defined, and if it is defined it will be used. Some people say that not having a difference between + and += prevents errors, but isn't this my own problem?

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A better design method is Explicit Casting. You can definitely overload Casting.


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