8

I am writing a C# class that is exposed to VB.Net. I would like to overload the vb.net ^ operator so that I can write:

Dim c as MyClass
Set c = New ...
Dim d as MyClass
Set d = c^2

In C#, the ^ operator is the xor operator and the power operator doesn't exist. Is there a way I can do this anyway?

D Stanley
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edeboursetty
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  • Not sure if this works in vb.net from a C# lib, but here's a similar post: http://stackoverflow.com/questions/8069665/or-xor-and-overloading – DougEC Nov 06 '12 at 20:55

2 Answers2

12

EDIT

It turns out there's a SpecialNameAttribute that lets you declare "special" functions in C# that will allow you (among other things) to overload the VB power operator:

public class ExponentClass
{
    public double Value { get; set; }

    [System.Runtime.CompilerServices.SpecialName]
    public static ExponentClass op_Exponent(ExponentClass o1, ExponentClass o2)
    {
        return new ExponentClass { Value = Math.Pow(o1.Value, o2.Value) };
    }
}

The op_Exponent function in the class above gets translated by VB into the ^ power operator.

Interestingly, the documentation states the Attribute in not currently used by the .NET framework...

--ORIGINAL ANSWER --

No. The power (^) operator gets compiled as Math.Pow() so there's no way to "overload" it in C#.

From LinqPad:

Sub Main
    Dim i as Integer
    Dim j as Integer
    j = Integer.Parse("6")
    i = (5^j)
    i.Dump()
End Sub

IL:

IL_0001:  ldstr       "6"
IL_0006:  call        System.Int32.Parse
IL_000B:  stloc.1     
IL_000C:  ldc.r8      00 00 00 00 00 00 14 40 
IL_0015:  ldloc.1     
IL_0016:  conv.r8     
IL_0017:  call        System.Math.Pow
IL_001C:  call        System.Math.Round
IL_0021:  conv.ovf.i4 
IL_0022:  stloc.0     
IL_0023:  ldloc.0     
IL_0024:  call        LINQPad.Extensions.Dump
D Stanley
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  • +1. That's what I expected. Just could not find a proof link. – Victor Zakharov Nov 06 '12 at 20:59
  • Mmmh, but VB.Net allows for the ^ operator overloading, and that compiles into a dll that can be loaded somewhere else. So the ^ operator overloading must be some valid IL. The question is, can this IL be generated from C#... – edeboursetty Nov 06 '12 at 21:02
  • @d--b: following same principle, it probably gets compiled to call whatever your implementation is, rather than storing it as an operator override. – Victor Zakharov Nov 06 '12 at 21:04
  • @Neolisk: but how could the compiler know what to do? If I create a class say `Matrix` in VB.Net with an overload on ^ and put it in a dll (Matrix.dll). How will the compiler know **just by loading the dll** that the operator has been overloaded on that class? – edeboursetty Nov 06 '12 at 21:11
  • @d--b: I just tried - it appears that `^` operator gets translated into `op_Exponent` function, so this is what's available to you from C#. – Victor Zakharov Nov 06 '12 at 21:25
  • The overload gets compiled to a static `op_Exponent` function that is callable by that name from C#. I have yet to find a way to recreate this operator from C# such that it will be recognized by VB. – D Stanley Nov 06 '12 at 21:26
  • @d--b I have found the secret attribute that allows you to overload the operator. See my edited answer. – D Stanley Nov 06 '12 at 21:39
  • @DStanley: Amazing! Can I somehow give you another upvote for finding that? – Victor Zakharov Nov 06 '12 at 23:53
  • Interestingly enough, even though F# emits the name `op_Exponentiation` for the exponentiation operator (`**`), this trick doesn't work for F#. Instead, F# looks for a static method named `Pow`. What a mess. – Miroslav Policki Aug 15 '20 at 08:28
2

By experiment, it turns out that operator overloading is just syntactic sugar, and better be avoided, if you need to develop in multiple languages. For example, ^ operator of VB.NET gets translated into op_Exponent function, so this is what's available to you from C#.

Why doesn't C# have a power operator?

You can use a native .NET way so you don't rely on operators:

Math.Pow(x, y);

Also for y=2 it is faster to use multiplication (x*x).

Victor Zakharov
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  • @Joey: could you please elaborate on how you would replace power with addition, and make it faster than multiplication? Not quite apparent to me. – Victor Zakharov Nov 06 '12 at 21:03
  • Ah, nevermind; I was thinking of 2 ⋅ *x* being often slower than *x* + *x*. Had my mind mixed up for a moment there. – Joey Nov 06 '12 at 21:41
  • @Joey: no problem, and thanks for pointing at a typo in my answer - now fixed. – Victor Zakharov Nov 06 '12 at 23:57