C: Accessing pointer to pointer to struct element from pointer to structure

Question

I want to access members of a struct from double pointer but I get the error

"error: expected identifier before ‘(’ token"

• C Double Pointer to Structure
• double pointer to struct inside struct

:

struct test{
  struct foo **val;
};

struct foo{
  int a;
}

int main (){
  struct test *ptr = (struct test *)malloc(sizeof(struct test));
  ptr->val = &foo;
  /*foo is already malloced and populated*/
  printf ("Value of a is %d", ptr->(*val)->a);
}

I've also tried:

*ptr.(**foo).a

Answer 1

You want to do this:

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

struct test {
    struct foo **val;
};

struct foo {
    int a;
};

int main(void) {
    struct test* test_ptr = malloc(sizeof(struct test));
    struct foo* foo_ptr = malloc(sizeof(struct foo));
    foo_ptr->a = 5;    // equivalent to (*foo_ptr).a = 5;
    test_ptr->val = &foo_ptr;
    printf ("Value of a is %d\n", (*(test_ptr->val))->a);
    free(test_ptr);
    free(foo_ptr);
    return 0;
}

Output:

C02QT2UBFVH6-lm:~ gsamaras$ gcc -Wall main.c 
C02QT2UBFVH6-lm:~ gsamaras$ ./a.out
Value of a is 5

In my example:

1. I allocate dynamically space for a struct test.
2. I allocate dynamically space for a struct foo.
3. I assign the value 5 to the member a of foo_ptr.
4. I assign the address of the allocated object of struct foo to the member val of test_ptr.
5. I print member a the struct that the double pointer val points to.

---

Note, in your example: struct foo is a type, so it doesn't make sense to ask for its address.

Also, you were missing a semicolon when you were done with the declaration of struct foo.

Oh, and make sure not to cast the return value of malloc().

Answer 2

in ptr->val = &foo;, foo is a struct (you declared it in lines 5 to 7). Taking its address does not give a **, but only a *.

Also it seems multiple things have the same names; is a foo the name of a structure or an instance of it, or both?

Then, when you dereference it: ptr->(*val)->a does seem the wrong sequence. as ptr->val is the address of foo (that's what you assigned in the line above it), what would ptr->(*val) be??

I think ptr->val.a would give you your a. But still, the val is declared as a ** and consistently used as a *. It might work, but makes not much sense.