Parameter passing multiple values using void pointer

Question

I want to pass multiple arguments to a function using a void pointer.

void* function(void *params)
{
//casting pointers
//doing something
}
int main()
{
  int a = 0
  int b = 10;
  char x = 'S';
  void function(???);
  return 0;
}

I know that I have to cast them to a certain variable in my function but I do not know how I can pass my 3 arguments as one void pointer to my function.

I have searched for this problem know quite some time but I could not find anything that would help me.

Create a struct with whatever members you need and pass a pointer to that. — Mat, Oct 15 '18 at 07:50
Good idea I am currently using two structs. One like this struct {void *value1; void * value2}params. And one struct that has two integers how can I pass my second struct in my first void *value struct? — Mark, Oct 15 '18 at 07:56
@StoryTeller: because the question isn't clear enough to answer. Why a `void*`? Are the arguments fixed or is the OP looking for `varargs`? Or some something else entirely (e.g. try to inspect the caller's stack or something weird like that)? — Mat, Oct 15 '18 at 08:00
@Mat - Then ask for clarifications. Honestly, that reads like an excuse. — StoryTeller - Unslander Monica, Oct 15 '18 at 08:02
@Mark: Please add clarifications to the question itself as an update. There is an "edit"-button below the question. — alk, Oct 15 '18 at 08:04

alk · Accepted Answer · 2018-10-15T08:23:24.747

You could do it like this:

struct my_struct
{
   int a;
   int b;
   char x;
}

void * function(void * pv)
{
  struct my_strcut * ps = pv; /* Implicitly converting the void-pointer 
                              /* passed in to a pointer to a struct. */

  /* Use ps->a, ps->b and ps->x here. */

  return ...; /* NULL or any pointer value valid outside this function */
}

Use it like this

int main(void)
{
  struct my_struct s = {42, -1, 'A'};

  void * pv = function(&s);
}

Following up on the OP's update:

struct my_struct_foo
{
   void * pv1;
   void * pv2;
}

struct my_struct_bar
{
   int a;
   int b;
}

void * function(void * pv)
{
  struct my_strcut_foo * ps_foo = pv; 
  struct my_struct_bar * ps_bar = ps_foo->pv1;

  /* Use ps_foo->..., ps_bar->... here. */

  return ...; /* NULL or any pointer value valid outside this function */
}

Use it like this

int main(void)
{
  struct my_struct_bar s_bar = {42, -1};
  struct my_struct_foo s_foo = {&s_bar, NULL};

  void * pv = function(&s_foo);
}

Thank you very much alk that was exactly what I was looking for. Now I understand what I have to do. — Mark, Oct 15 '18 at 08:10

Aif · Answer 2 · 2018-10-15T08:07:08.687

1

The void* is used as a pointer to a "generic" type. Hence, you need to create a wrapping type, ~~cast~~ convert to void* to invoke the function, and ~~cast~~ convert back to your type in the function's body.

#include <stdio.h>

struct args { int a, b; char X; };
void function(void *params)
{
  struct args *arg = params;
  printf("%d\n", arg->b);
}
int main()
{
  struct args prm;
  prm.a = 0;
  prm.b = 10;
  prm.X = 'S';
  function(&prm);
  return 0;
}

edited Oct 15 '18 at 08:07

answered Oct 15 '18 at 08:00

Aif

11,015
1
30
44

1

Nitpicking: Your code does not cast at all. What happens is called "implicit conversion". – alk Oct 15 '18 at 08:02
I thought type conversion were more like integer promotion or boolean conversions.. (?) – Aif Oct 15 '18 at 08:05
In C there are two kinds of converting types: "Implicit Conversion" (promotion is one of this) and "Explicit Conversion", which is applied using a casting operation. Here `int i = 42; double d = ((double) i) / 7;` `i` is converted explicitly (to `double`) and `7` implicitly (to `double`) via promotion. – alk Oct 15 '18 at 08:17

Parameter passing multiple values using void pointer

2 Answers2