Reputation: 906
I was digging into a 3rd party code base and found that it is apparently valid to declare a type as a pointer to an undefined struct. As a minimum working example, consider a C file test.c
containing nothing but:
typedef struct foo *bar;
What surprises me is that this file compiles without any problems using the command
gcc test.c -shared
Why does the compiler not complain about the struct foo not being declared anywhere?
My environment is Ubuntu 16.04 with gcc (Ubuntu 5.4.0-6ubuntu1~16.04.5) 5.4.0 20160609.
Upvotes: 3
Views: 477
Reputation: 223872
The declaration above creates a forward declaration of struct foo
. Although you can't access its members, you can operate on a pointer to it.
This is commonly referred to as an opaque type, and is used to hide implementation details of a library from users of the library.
For example a library implementation may contain the following:
lib.c:
struct foo {
int f1;
};
struct foo *init()
{
return malloc(sizeof(struct foo));
}
void set1(struct foo *p, int val)
{
p->f1 = val;
}
int get1(struct foo *p)
{
return p->f1;
}
void cleanup(struct foo *p)
{
free(p);
}
The header file for this library might look like this:
lib.h:
struct foo;
struct foo *init(void);
void set1(struct foo *p, int val);
int get1(struct foo *p);
void cleanup(struct foo *p);
The user of this library would use the init
function to create an instance of the struct and the set1
and get1
functions to read / update the member. The user can't however create an instance of struct foo
or access the members without going through one of the interface functions.
Upvotes: 4
Reputation: 30926
Because pointer to structs have same size (for any structs). The size of a structure pointer doesn't depend on it's definition. Because no matter what, the pointer to it would be behaved the same way as any other structures pointer.
Upvotes: 2