Create a pointer of the type (pointer to struct) and assign the address of an instance of the structure to your pointer:
typedef struct x { /* ... */ };
struct x my_structure;
struct x* ptr = &my_structure;
Not in C, no.
It is a pointer that points to a member of a structure.
struct thisorthat *ptr;
by using structure in c.........
Nothing special: struct foo { void *ptr; };
The C language does not support references, that is, the C++ concept of creating an alias to a variable. You can create pointers and dereference them, or you can use the preprocessor's #define mechanism to use another name for a variable. Even in C++, however, you cannot create a pointer to a reference. If you try to, you will end up creating a pointer to the original value. This is because all a reference is is an automatically dereferenced pointer - a compiler shortcut, rather than a totally new feature. So the limitation of not being able to point at a reference isn't really too bad.
Pointer, eg: struct Person { struct Person *father, *mother, *spouse; int age; char *name; };
A pointer is a variable that holds address information. For example, in C++, say you have a Car class and another class that can access Car. Then, declaring Car *car1 =new Car() creates a pointer to a Car object.. The variable "car1" holds an address location.
Not really, but you can have: - a pointer pointing to a structure (FILE * is an example) - a pointer pointing to a structure-member (eg: struct tm tm; int *ip= &tm.tm_year) - a structure-member that is a pointer (any type) Example: typedef struct TreeNode { struct TreeNode *left, *right; int data; } TreeNode; TreeNode *root = (TreeNode *)calloc (sizeof (TreeNode), 1);
Yes. If the ports are memory mapped, then you simply need a pointer to that address, and you need to declare the pointer as volatile. If they are I/O mapped, then you need to create an _asm{} block.
Increment or decrement the pointer by the required offset.
no