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From: "rb" <rbinion AT rcn DOT com>
Newsgroups: comp.lang.c++,comp.lang.c++.moderated,comp.os.msdos.djgpp
Subject: Re: Heterogenous object container without templates & type casting ???
Date: 13 Feb 2002 01:28:48 -0500
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To: djgpp AT delorie DOT com
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Reply-To: djgpp AT delorie DOT com

Well technically speaking since C is roughly a subset of C++ your code is
already "transformed".

First I would ask why you don't want to use templates? Templates are
perfectly suited for what you want to accomplish.

Without templates I'm afraid you are "stuck" with the casts. In C and C++ a
pointer to anything is a void* which you obviously know. As for the rest of
your code, are you trying to win an obsfucated C++ contest? :)

I'm kidding, but here are a couple of comments about the code you posted.

1. typedefs for structs are not needed in C++.
2. "this" is a keyword in C++, so you can't use it as a variable name.
3. Maybe this is your question and I am just stating the obvious but you
have to change your 3 functions from this

> int*    get2(A* this) {return this->_p;}
> double* get3(A* this) {return this->_p;}
> char*   get4(A* this) {return this->_p;}

to this.

int*    get2(A* that) {return (int*)that->_p;}
double* get3(A* that) {return (double*)that->_p;}
char*   get4(A* that) {return (char*)that->_p;}

In C++ you can't assign a void* to a non void*  without doing an explicit
cast.

4. The agregate assignment of your four variables A,B,C,D are probably
handled better by constructors.

5. I am probably missing the overall gist of what you are trying to do. But
if you want the _p member of struct A to be accesible as an int, double and
char .. wouldn't you be better of just using a union?

6. Finaly here is roughly the equivalent C++ code with templates. Note I
used the c io routines rather than the C++ iostreams so as not to further
complicate the issue, made everything public for the same reason, and also
left out other C++ idioms (like using const, ctor's, etc.).

template<typename T> class holder
{
public:
    T value_;
};

int main(int argc,char* argv[])
{
 int value = 66;

 holder<int> b = {value};
 holder<double> c = {value};
 holder<char> d = {value};

 printf("I am of type \"int\". My value is %i\n",(b.value_));
printf("I am of type \"double\". My value is %f\n",(c.value_));
printf("I am of type \"char\". My value is %c\n",(d.value_));

 return 0;
}


Hope that helps,
Bob



"Traveler" <traveler AT netti DOT fi> wrote in message
news:a48ggn$f8n$1 AT tron DOT sci DOT fi...
> Hi
>
> Is it possible to transform the following C code to C++ so that there is
no
> need for type-casting ?
> I am trying to make a general object class that can contain any type
without
> the use of templates.
>
>
> #include <stdio.h>
> #include <stdlib.h>
>
> typedef struct  A
> {
>         void*   _p;
>         void* (*get)(struct A* this);
> }     A;
>
> typedef struct
> {
>         int*  (*get)(A* this);
> } B;
>
>
> typedef struct
> {
>         double* (*get)(A* this);
> } C;
>
> typedef struct
> {
>         char*   (*get)(A* this);
> } D;
>
> void*   get1(A* this) {return this->_p;}
> int*    get2(A* this) {return this->_p;}
> double* get3(A* this) {return this->_p;}
> char*   get4(A* this) {return this->_p;}
>
> int     main(int argc,char* argv[])
> {
>         int     value = 66;
>         A       a = {&value,get1};    // Simple & dangerous way to "put"
> something in to void pointer "a._p"
>                                                     // because I had no
> patience to play games with "malloc" in this case.......
>         B       b = {get2};            // Takes care of the type
casting....
>         C       c = {get3};
>         D       d = {get4};
>
>         printf("I am of type \"int\". My value is %i\n",*b.get(&a));
>         printf("I am of type \"double\". My value is %d\n",*c.get(&a));
>         printf("I am of type \"char\". My value is %c\n",*d.get(&a));
>
>         return(0);
> }

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