How do you declare a pointer to a structure in C?

Answer 1

It depends on whether you've declared a type alias or not. Without an alias, you must include the struct keyword:

struct record {

// ...

};

struct record* ptr; // declare a pointer to struct record

With an alias, the struct keyword can be omitted from the type name. By convention, we use the _t suffix to denote the type name. Typically, aliases are declared as part of the type declaration:

typedef struct record_t { // type name

// ...

} record; // alias

record* p; // declare a pointer to record

Note that record is the alias while struct record_t is the data type associated with that alias. We can use the type or the alias as we see fit, however type aliases are more convenient.

We can also declare pointer aliases in the type declaration:

typedef struct record_t {

// ...

} record, *record_ptr;

record_ptr p; // declare a pointer to record

However, this latter technique is best avoided because we can often end up inadvertently declaring pointer to pointer types instead of pointers:

record_ptr* p; // equivalent to: record** p; -- possibly erroneous

Note that you cannot use a type alias within a self-referential data type, such as a node with embedded pointers to other nodes (as used in linked list implementations):

typedef struct node_t {

int data;

node* next; // error: node is not (yet) defined

node* prev; // error: node is not (yet) defined

} node;

To fix these errors, use the actual type name:

typedef struct node_t {

int data;

struct node_t* next;

struct node_t* prev;

} node;

Answer 2

typedef int (*myfunptr)(int argc, char **argv);

myfunptr m= main;