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

myfun[string]; //calls the overloaded [] operator;


lotsofunptr->[string];//error

help is much apreciated

matt p wrote:
> example:
> FunClass myfun;
> FunClass *lotsofunptr=&myfun;
>
> 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 wrote:
> matt p wrote:
>
>> example:
>> FunClass myfun; FunClass *lotsofunptr=&myfun;
>>
>> 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.'

matt p wrote:

> example:
> FunClass myfun;
> FunClass *lotsofunptr=&myfun;
>
> myfun[string]; //calls the overloaded [] operator;
>
>
> lotsofunptr->[string];//error

(* lotsofunptr) [string];

lotsofunptr->operator [] (string);

--
Salu2