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

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:
I allocate dynamically space for a struct test.
I allocate dynamically space for a struct foo.
I assign the value 5 to the member a of foo_ptr.
I assign the address of the allocated object of struct foo to the
member val of test_ptr.
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.