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operator overloading

Started by Freaker13 Oct 14, 2004 at 7:28 AM 6 replies 1.2k views

Original Post

Oct 14, 2004 at 7:28 AM

Hi all, I've oveloaded the -> operator in my VTcMemSlot class but when i do slot->getPtr the overloaded operator isn't called.(slot is of type VTcRAMSlot which inherit from VTcMemSlot)


class VTcMemSlot
{
	friend class VTcCacheManager;
	friend class VTcCacheSlotOwner;
	
public:
	VTcMemSlot(VTcCacheManager*	manager, unsigned int bytes);
	~VTcMemSlot();

	inline VTcMemSlot* VTcMemSlot::operator ->()
	{
		manager->logger->wrStr(1,"operator->");
		if(prev == NULL)
			manager->cacheListsLRU[indexInCacheList] = next;//zet de tweede op de eerste plaats
		else
		{
			//link prev and next
			prev->next = next;
			next->prev = prev;
		}
		//move slot to end of linked  list
		manager->lastUsedSlots[indexInCacheList]->next = this;
		prev = manager->lastUsedSlots[indexInCacheList];
		next = NULL;
		manager->lastUsedSlots[indexInCacheList] = this;
		return this;
	}
	
	void	splitSlot		();
	void	setSizeInBytes	(unsigned int bytes){this->bytes = bytes;}

	virtual void*	getPtr() = 0;

thx in advance

RabidCow

Oct 14, 2004 at 9:30 AM

It would be helpful to see the code in which you call the overloaded operator. That aside, I'll explain what I think the problem might be.

Overloaded operators only work when you call them on the type they are defined for. For instance, if you define:

A::operator->()

then this won't call your operator:

A *a;
a->whatever

because a is a pointer, and not an object of the class A. In order for it to work, you need to call it like this:

A a;
a->whatever

I don't know which one you meant to call in VTcMemSlot::getSizeInBytes, but since this is a pointer, you're not going to be able to call the overloaded -> operator on it unless you dereference it first.

There. Clear as mud?

-RC

Freaker13

Oct 14, 2004 at 3:01 PM

thx that's the solution. Didn't know you could do -> on an obj.

x_gamer_x

Oct 14, 2004 at 8:57 PM

You can, so long as you overload it ;).

Tis' the beauty of overloading.

raptorstrike

Oct 14, 2004 at 9:07 PM

just to examine this further what if you really did need to call it from a pointer would you do something like this?

a->b->c? or would you need to change the way you set it up competely?

silvermace

Oct 14, 2004 at 9:23 PM

if you wanted to call it from a pointer, you'd dereference
eg.

A *p;(*p)-> // this will call the overloaded operatorp-> // this will call the default (pointer) access operator

get it?

_Phalanx_

Oct 15, 2004 at 12:24 AM

i dont get whats so great about operator overloading. Wouldnt it be just as easy to write a funtion or macro that has 2 parameters? That way people looking at your code would realize whats happening instead of being mystified.

silvermace

Oct 15, 2004 at 1:04 AM

Quote:
Original post by _Phalanx_
i dont get whats so great about operator overloading. Wouldnt it be just as easy to write a funtion or macro that has 2 parameters? That way people looking at your code would realize whats happening instead of being mystified.

one way to look at it, its just a convenience feature really :)
in most situations using an operator can make things clearer

a.setValue( b.objPtr->next.getValue() + c.getValue() )

could become
a = b->next+ c

isnt that a wee bit easier to read?

PS. thats BS code, but get the picture?

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