Calling a object's overloaded operator with a pointer to that object.

[matt p]

example:
FunClass myfun;
FunClass *lotsofunptr=&m yfun;

myfun[string]; //calls the overloaded [] operator;
lotsofunptr->[string];//error

help is much apreciated

[Karthik Kumar]

matt p wrote:

example:
FunClass myfun;
FunClass *lotsofunptr=&m yfun;

myfun[string]; //calls the overloaded [] operator;
How does your signature of the overloaded function look like ?
And what is 'string' . C++ std. specifies it to be a type in std
namespace.


lotsofunptr->[string];//error

Post compilable code here to seek help.

--
Karthik. http://akktech.blogspot.com .
'Remove _nospamplz from my email to mail me.'

[Karthik Kumar]

Karthik Kumar wrote:

matt p wrote:

example:
FunClass myfun; FunClass *lotsofunptr=&m yfun;

myfun[string]; //calls the overloaded [] operator;
Assuming string to be a variable, (a bad choice for naming it
though) and of the same type as the overloaded function would expect ,
here it goes.

myfun[string] ;

is essentially

myfun.operator[](string)

How does your signature of the overloaded function look like ?
And what is 'string' . C++ std. specifies it to be a type in std
namespace.

lotsofunptr->[string];//error

So if you want to get the same thing as that of a pointer , use

(*lotsofunptr)[string];

You essentially dereference the pointer and apply the same syntax.
If you are not happy then use

lotofunptr->operator[](string)

That should work fine too.

Post compilable code here to seek help.

--
Karthik. http://akktech.blogspot.com .
'Remove _nospamplz from my email to mail me.'

[Julián Albo]

matt p wrote:

example:
FunClass myfun;
FunClass *lotsofunptr=&m yfun;

myfun[string]; //calls the overloaded [] operator;
lotsofunptr->[string];//error

(* lotsofunptr) [string];

lotsofunptr->operator [] (string);

--
Salu2