linked lists .

This is a discussion on linked lists . within the C Programming forums, part of the General Programming Boards category; PLZ help ! the file attached . i dont understand the idea of this function Code: void reg_student(slist *students,clist *courses,int ...

  1. #1
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    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:
    void reg_student(slist *students,clist *courses,int id, int number)
    student->course = newCourse ->A->B->C->NULL ;
    Attached Images Attached Images

  2. #2
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    יפה, בדיוק אותה שאלה לשנינו
    מי זה??

  3. #3
    ATH0 quzah's Avatar
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    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.
    Hope is the first step on the road to disappointment.

  4. #4
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    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.

  5. #5
    C++ Witch laserlight's Avatar
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    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).
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