Storing arbitrary objects in array in C

问题

I am new to C but am trying to wrap my head around trying to store arbitrary objects in an array. Structs, integers, chars, functions, etc. Basically something perhaps using void pointers along the lines of (pseudocode):

void *array[] = malloc(10000);
struct MyStruct m = malloc(sizeof(m));
int x = 10;
char c[] = "Look Here";
array[0] = &m;
array[1] = &x;
array[2] = &c;

Essentially I want to have a global array store arbitrary objects sort of like a database, and then fetch them by index somehow.

void *global_array[];

void
get_from_array(int index, void *ptr) {
  *ptr = global_array[index];
}

int
main() {
  global_array = malloc(10000);
  struct MyStruct m = malloc(sizeof(m));
  int x = 10;
  char c[] = "Look Here";
  global_array[0] = &m;
  global_array[1] = &x;
  global_array[2] = &c;
  struct MyStruct m2;
  get_from_array(0, &m2);
  assert(m == m2);
}

Is anything like this possible?

回答1:

Yes. You can create a void double pointer void** And allocate it space of (say 10000) void pointers with malloc. It can be indexed and it effectively acts as an array of void* type

For the code mentioned in your question, it would be something like

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


void **array;

typedef struct MyStruct{
  int a;
  char b;
}MyStruct;

int main()
{
  array = malloc(sizeof(void*)*10000);
  struct MyStruct* m = (MyStruct*)malloc(sizeof(MyStruct));
  m->a=1;
  m->b='x';
  int x = 10;
  char c[] = "Look Here";
  array[0] = m;
  array[1] = &x;
  array[2] = &c;
  printf("%d %c\n%d\n%s\n",((MyStruct*)(array[0]))->a,((MyStruct*)(array[0]))->b,*(int*)(array[1]),(char*)(array[2]));
  return 0;
}

回答2:

If you want to store many types, you need to save the size or the type of each variable. You should use the struct and union. For example:

typedef enum EltType { TYPE_STRING, TYPE_INT, TYPE_FLOAT } TYPE;
typedef struct Element {
  TYPE type;
  union {
    char  *str;
    int    i;
    float  f;
  };
}ELEMENT;

The test:

#include <stdio.h>

typedef enum EltType { TYPE_STRING, TYPE_INT, TYPE_FLOAT } TYPE;
typedef struct Element {
  TYPE type;
  union {
    char  *str;
    int    i;
    float  f;
  };
}ELEMENT;

void print_value(ELEMENT elt) {
    switch (elt.type) {
        case TYPE_STRING:
           printf("%s\n", elt.str);
           break;
        case TYPE_INT:
           printf("%d\n", elt.i);
           break;
        case TYPE_FLOAT:
           printf("%f\n", elt.f);
           break;
    }
}
int main(int argc, char const *argv[])
{
    ELEMENT elt1, elt2, elt3; 
    elt1.type = TYPE_STRING;
    elt1.str = "string";

    elt2.type = TYPE_INT;
    elt2.i = 5;

    elt3.type = TYPE_FLOAT;
    elt3.f = 1.2;

    print_value(elt1);
    print_value(elt2);
    print_value(elt3);

    return 0;
}

来源：https://stackoverflow.com/questions/61156025/storing-arbitrary-objects-in-array-in-c

标签

arrays

database

pointers

void-pointers