passing operators?

**m37h0d** · 05-13-2008

i'm guessing this is impossible, but is it possible to point to an operator's function, i would ideally like to pass it as an argument.

i would like to avoid having to pass a string literal to represent the operator and writing an if statement for each operator.

is there any way one could achieve something akin to:

Code:

template<typename returnType,typename Arg>returnType func(Arg arg1,Arg arg2,operator op)
{
}
matrix a;
matrix b;
func(a,b,a.*);

there are probably errors in the above, it is meant only to illustrate the concept.

it seems like there should be some way to create a function pointer to the operator, but i cannot get it...

it would be nice if it were possible to do this for POD also.

**cpjust** · 05-13-2008

Look in the <functional> header.
It has lots of templates for things like:
plus<>
minus<>
multiplies<>
divides<>
...

**m37h0d** · 05-13-2008

templatizing the operation wasn't the issue; i need to store it!

is there a way i can pass one of those as an arg?

currently i have a series of if statements and corresponding member functions in the template class, but it would be nice if i could make it more eloquent.

**Elysia** · 05-13-2008

How about just...

Code:

template<typename T> void foo(T arg) { }
int main()
{
    obj myobj;
    foo(&myobj::operator+);
    // Or foo(&myobj.operator+);
}

Tried it?

**cpjust** · 05-13-2008

Originally Posted by m37h0d

templatizing the operation wasn't the issue; i need to store it!

is there a way i can pass one of those as an arg?

currently i have a series of if statements and corresponding member functions in the template class, but it would be nice if i could make it more eloquent.

Most of those derive from binary_function<Type, Type, Type>, so you could probably store the function object as a binary_function<Type, Type, Type>& in your code...

**m37h0d** · 05-13-2008

@ Elysia; indeed, i had considered that, but what if T is an int or other POD type?

**m37h0d** · 05-13-2008

Originally Posted by cpjust

Most of those derive from binary_function<Type, Type, Type>, so you could probably store the function object as a binary_function<Type, Type, Type>& in your code...

i will look into that. thank you.

**Elysia** · 05-13-2008

Of course, silly me, it must be:

Code:

template<typename T> void foo(T arg)
{ }

class obj
{
public:
	void operator + (int) {}
};
int main()
{
	obj myobj;
	foo(&obj::operator+);
}

Works for me.
Now you can call your operator from the template function.

**CornedBee** · 05-13-2008

But that limits you to member +. It already doesn't work for a free operator +.

You could just store a std::plus in a boost::function or a std::tr1::function.

**Elysia** · 05-13-2008

Originally Posted by m37h0d

@ Elysia; indeed, i had considered that, but what if T is an int or other POD type?

I don't see why you can't simply pass the address of an appropriate operator function. Even if it's free, you should be able to call it and pass the arguments.

Code:

template<typename T, typename T2> void foo(T arg, T2& arg2)
{ 
	arg(arg2, 0);
}

class obj
{
public:
	friend void operator + (const obj&, int);
};

void operator + (const obj&, int) {}

int main()
{
	obj myobj;
	foo(&operator+, myobj);
}

**CornedBee** · 05-13-2008

I don't see why you can't simply pass the address of an appropriate operator function.

Built-in operators aren't functions.

**Elysia** · 05-13-2008

Ah, you mean built-in ones. I suppose this would only work on overloaded operators.

**m37h0d** · 05-13-2008

Originally Posted by Elysia

I don't see why you can't simply pass the address of an appropriate operator function. Even if it's free, you should be able to call it and pass the arguments.

Code:

template<typename T, typename T2> void foo(T arg, T2& arg2)
{ 
	arg(arg2, 0);
}

class obj
{
public:
	friend void operator + (const obj&, int);
};

void operator + (const obj&, int) {}

int main()
{
	obj myobj;
	foo(&operator+, myobj);
}

that can't possibly work. how could the template know that the type arg is a function?

**Elysia** · 05-13-2008

The compiler deduces the type depending on what you pass.
So what you pass IS the type, so you can use it AS that type.
That's the power of templates.
It does work.

**m37h0d** · 05-13-2008

perhaps i am being dense in not spotting a trivial error there, but that does not compile.

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