CODEWITHSUNDEEP
cstructtypedef
Granolaboy
Granolaboy

Reputation: 333

C struct inside a struct

I have a typedef structure which defines an element in a queue. Inside the definition I have another typedef structure with its own values.

typedef struct q_elem_s
{
    def_task task;
    int arrive_time;
    struct q_elem_s *next;
} q_elem;

Inside q_elem_s is def_task task defined by:

typedef struct task_s
{
    int id;
    int length;
} def_task;

When I want to access a value inside def_task task I get following error:

"invalid type argument of '->' (have 'def_task')

What am I doing wrong?

Upvotes: 0

Views: 1246

Answers (2)

Paulo Bu
Paulo Bu

Reputation: 29794

Both -> and . operator are used to access structure fields. The main difference is that -> is used when you have a pointer to a structure variable and . is used when you have a structure variable. For example:

q_elem* q;   /* q stores the address to a q_elem structure */
q_elem q;    /* q stores the q_elem structure */

Now, in the first case, to access structure's members you'll have to refer them by using -> operator.

q_elem* q;
q->task;     /* access to def_task */

If you instead have a structure variable you can do:

q_elem q;
q.task;     /* access to def_task */

Inside your structure q_elem, task is defined as a structure and not as a pointer to structure. This means that once you have access to the task structure inside q_elem, accessing task's fields is achieved by . (dot) operator.

This leave you with two choices:

q_elem* q;
q_elem->task.id

or

q_elem q;
q_elem.task.id

Notice the difference between the q_elem variable declaration.

I hope this explain things a little.

Upvotes: 3

Paul Rubel
Paul Rubel

Reputation: 27222

If you have a q_elem and you want to access id from its task you'd use task.id. Note that you don't have a def_task pointer (def_task*), which would need an arrow but rather a def_task value. Try doing something like so:

q_elem q;
q.task.id = 12;

Upvotes: 2

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