linked lists .

PLZ help !

the file attached .

i dont understand the idea of this function

Code:

---

void reg_student(slist *students,clist *courses,int id, int number)

---

this function is VOID , how can add course to student by void function ;

how to change student->course ?

supose student->course = A->B->C->NULL ;

can i had a newCourse ?

after using

Code:

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void reg_student(slist *students,clist *courses,int id, int number)

---

student->course = newCourse ->A->B->C->NULL ;

יפה, בדיוק אותה שאלה לשנינו :)
מי זה??

English please.

All functions take values. The first two values this function takes are memory addresses to structures. Those structures will also have pointer. So, let's assume it's like so:

Code:

---

struct foo
{
    struct foo *next;
    ...data...
};

---

Then we make an instance of that structure, and set its pointer to point to NULL.

Code:

---

struct foo *instance = malloc( sizeof *instance );
instance->next = NULL;

---

Now, if I pass the address of 'instance' to a function, I can access 'instance->foo' inside that function, because, through the address, I can access all of its members. Since I can access that member, I can make that member's value change--and in this case, I can make it point to a new structure.

Code:

---

void bar( struct foo *p )
{
    if( p )
    {
        struct foo *q = malloc( sizeof *q );
        if( q )
        {
            q->next = NULL;
            p->next = q;
        }
    }
}

---

Assuming of course, that we know the structure we are passing has a NULL for its 'next' member. Otherwise you'd naturally want to tweak this a bit to handle if it were not null all ready.


Quzah.

Code:

---

void func1( int i ){} //pass by reference  (can't manipulate contents unless i is global
void func2( int *i ){} //pass by address (manipulate contents of non-local variable

int i;

func1( i );
func2(&i)

// &i gives the address, pasising a pointer to it in memory.

---

if I didn't use int i, and wanted to use int *i to innitialize it, I would be responsible for allocating memory sizeof(int) for it to hold a value.

"->" is just that, a pointer to a structure's member in memory, if you have multiple instances of a struct you must always use ->, otherwise you can use a "." to access its members.

Just a few minor corrections with respect to tempster09's post #4:

Code:

---

void func1( int i ){} //pass by reference  (can't manipulate contents unless i is global

---

The above actually demonstrates pass by value, not pass by reference.

Code:

---

void func2( int *i ){} //pass by address (manipulate contents of non-local variable

---

It is true that an address is passed in the above function, and although the pointer is passed by value, this simulates pass by reference.

Quote:

---

Originally Posted by tempster09

"->" is just that, a pointer to a structure's member in memory, if you have multiple instances of a struct you must always use ->, otherwise you can use a "." to access its members.

---

I think tempster09 meant that -> is a form of pointer notation, and indeed a->b is syntactic sugar for (*a).b, where a is a pointer to a structure which has a member named b. The part about "multiple instances of a struct" sounds like a mistake: I think "a pointer to a struct" is the intended wording (although the "must always" clause is obviously not true).