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Function pointer and operator()?

Started by gautr Jul 12, 2006 at 8:23 AM 3 replies 650+ views

Original Post

Jul 12, 2006 at 8:23 AM

Hi there. I'd like to be able to pass classes with operator() to a function which takes other functions as arguments. Can't get it right though :(


typedef void (*FuncPtr)();

void DoStuffWithFuncPtr(FuncPtr p)
{
	p();
}

void SomeFunc()
{
	// Implementation
}

class CSomeClass
{
public:
	CSomeClass() {}
	~CSomeClass() {}

	void operator()() { /* Implementation */ }
};

void main()
{
	CSomeClass instance;

	// Doesn't work!
	DoStuffWithFuncPtr(instance);
	// "error C2664: 'DoStuffWithFuncPtr' : cannot convert parameter 1 from 'CSomeClass' to 'FuncPtr'
    //  No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called"

	// This naturally works though
	DoStuffWithFuncPtr(SomeFunc);
}

Any thoughts on how to overcome this?

SiCrane

Jul 12, 2006 at 8:26 AM

A class with an overloaded operator() is still a class and as such is not convertible to a function pointer type. You can create a template method that accepts both classes and function pointers such as:

template <typename Callable>void DoStuffWithFuncPtr(Callable p){	p();}

Alternately you can use something like boost::function to handle the abstraction between function pointers and function objects.

gautr

Jul 12, 2006 at 8:35 AM

Thanks :)

gautr

Jul 12, 2006 at 8:49 AM

Another question. How can I get the address operator() function in CSomeClass?

rip-off

Jul 12, 2006 at 8:53 AM

Quote:
Original post by gautr
Another question. How can I get the address operator() function in CSomeClass?

Member function pointers are not convertable to function pointers.

Typically a member function pointer will be bigger than the standard 32bit address on 32bit machines, as they have extra data in them.

But here you go anyway.

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