Re: void* passed as funtion parameters?

From:
=?Utf-8?B?Um9iYnk=?= <Robby@discussions.microsoft.com>
Newsgroups:
microsoft.public.vc.language
Date:
Wed, 2 Sep 2009 13:22:08 -0700
Message-ID:
<EE5F7855-14F4-4F64-AAEA-FA61A8592E21@microsoft.com>
Hello Uli,

Sorry for bothering you with the sample, that was very nice of you to take
time and get back with a sample code. It is very appreciated.

I have entered your solution in VC++ compiler to try it, but I do get the
following warning:

1>c:\_dts_programming\c_programming\c\c_tests\c_string_samples\misc_c_samples\enumpointer.c(59)
: warning C4715: 'get_f1' : not all control paths return a value

But before discussing the warning, I have a 2 questions and I will outline
them at the end of this post. In the mean time here is how I tried it, note I
typedefed my structs though:

================================

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

enum type {type_pc, type_ddlb};

// TABLES
typedef struct tag_pc_table
{
long z;
long h;
}pc_table;

typedef struct tag_lb_table
{
long z;
long u;
}lb_table;

// OBJECTS
typedef struct tag_pc {
long w;
struct tag_pc_table *dc;
enum type type;
} pc;

typedef struct tag_lb {
long w;
struct tag_lb_table *dc;
enum type type;
} lb;

void* get_f1(void* p)
{
enum type* t = p;

switch(*t)
{
  case type_pc:
  {
   struct pc* pc = p;
   return pc; //pc->f1;
  }
break;

case type_ddlb:
  {
   struct ddlb* ddlb = p;
   return ddlb; //ddlb->f1;
  }
break;
}
}

int main()
{
long t;
pc *a;
lb *b;

pc_table apc[] = {101, 201, 301, 302}; // Fill table
lb_table alb[] = {102, 202, 401, 402}; // Fill table

a = malloc(sizeof (struct tag_pc));
get_f1(a);

b = malloc(sizeof (struct tag_lb));
get_f1(b);

free(a);
free(b);
return 0;
}
=================================

1- One thing I don't get is when I pass in the "a" variable when calling the
get_f1() function, its address gets assigned to *t byt he following line: in
the get_f1() function:

enum type* t = p;

and then we put *t in the switch command/case command, but the type_pc and
type_ddlb are 0, and 1 respectively and really don't match with (*t) ?

2- Why are we returning pc->f1, I don't think f1 is a member of the
structures?

3- What are we supposed to return? >>> pc or ddlb?

Anyhow, I have been experimenting with this and it turns out that you guys
were right that it works without implicitly casting to the correct type. I am
still scratching my head on that one as to why it would still work. I didn't
use any templates, macro functions or base structs. I will post the example
in a new post... this one has pretty well got its usage! The new post will be
called "No casting of void*"

Uli, I still am currious about the questions I asked here!

Thanks for your input!

--
Best regards
Roberto

"Ulrich Eckhardt" wrote:

Robby wrote:

2. You could also attach the type to the struct itself, like in the first
member an enumeration. This would allow you to distinguish the type
easier. You could then treat the void pointer as pointer to the
enumeration, read it, and then treat the different types accordingly.


This seems interesting, but I don't quite understand what you mean by:

"You could also attach the type to the struct itself, like in the first
member an enumeration."

I have never done this. Is it possible to show a very short example...
perhaps 5 lines or so, I would really appreciate it. If its too much
trouble, then its okay too.


enum type {
  type_pc,
  type_ddlb,
};
struct pc {
  enum type type;
  ...
};
struct ddlb {
  enum type type;
  ...
};

void get_f1(void* p) {
  enum type* t = p;
  switch(*t) {
  case type_pc: {
      struct pc* pc = p;
      return pc->f1;
    }
    break;
  case type_ddlb: {
      struct ddlb* ddlb = p;
      return ddlb->f1;
    }
    break;
  default:
    assert(0);
  }
}

Uli

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