Sorry, can anyone explain to me how the offset [ ] operator works when it is overloaded?

  class YourClass{	NODETYPE thearray[somesize];public://...NODETYPE operator[](u32 Index){	// do some assertion check for in range	// ..	return(thearray[Index]);}};  

NODETYPE is whatever you want it to be.

Watch out for :

  YourClass inst1,inst2,*ptrinst=&inst1NODETYPE k = inst[3]; // correctNODETYPE i = ptrinst[3]; // wrong , indexing pointer array NODETYPE j= (*ptrinst)[3]; // correct, indexing YourClass instance.  

if you want to do :

inst1[3] = somevalue,

you need to past back a reference from the operator[]. then the operator= is called on the returned reference. But beware this allows things to muck about with the internal values. Better to use a Set( index, value ) function.

Note my comment at the bottom,

quote:
if you want to do :

inst1[3] = somevalue,

you need to past back a reference from the operator[]. then the operator= is called on the returned reference. But beware this allows things to muck about with the internal values. Better to use a Set( index, value ) function.

I left it out on purpose,

NODETYPE operator[](u32 Index ){}
Will return by copy constructor, meaning:-

u32 array[n];
u32 i = array[m];
i = 3;
Modifying i won't change the array contents.

But using
NODETYPE &operator[](u32 Index){}
Will return by reference, which in essance is:-

u32 array[n];
u32 *i = &array[m];
(*i) = 3;
so modifying contents of i directly modifies the contents of the array which may be unsafe,(after all a class is *supposed* to protect the data from this external fiddling!).

References are useful, but you have to be carful what gets access to them!


Edited by - Mark Duffill on February 26, 2002 4:05:36 AM