在 C 中将 * 变为 char 或 int

Question

I want to get any types of variables in my code, so I did a void * type to accept others. But I can get in char * but not in int values. And I don't understand how I can did it. Here my code :

void    insertion(t_liste *liste, void *newValue) {
  t_element *new = malloc(sizeof(void *));
  int i;
  int *j = &i;

  if (liste == NULL || new == NULL) {
    exit(EXIT_FAILURE);
  }
  if (newValue == j || (char *)newValue) {
     new->value = newValue;
     new->suivant = liste->premier;
     liste->premier = new;
     liste->taille++;
     new->index = liste->taille;
  }
}

In my main I did

insertion(maListe, 5);

it didn't work, but if I did this:

insertion(maListe, "test");

It works. But I want both works ! Here my .h

typedef struct s_element t_element;
typedef struct s_liste t_liste;

struct s_element{
  int           index;
  void          *value;
  t_element     *suivant;
  t_element     *precedent;
};

struct s_liste{
  t_element     *premier;
  t_element     *dernier;
  int           taille;
};

Any idea ?

Answer 1

OK! In your function void insertion(t_liste *liste, void *newValue) you are taking a argument of type void* . In the first case when you send a string(char *) the base address of the string is passed , so address is taken to newValue ,but in case when you pass a number,say 5 ,integer is passed to newValue where it expects an address.