• C

Strange problem with b-trees, and C functions in general

Subject: Strange problem with b-trees, and C functions in general

Hi to everyone...
I need your answer to a strange subject
that has to do with b-trees.
I want to built a function that will create a b-tree and will return its root node.
As parameters to the function I must pass the degree of the tree, the type of key and the key size.
For example:

CreateTree(root,degree,attrType,attrLength)

This function creats a tree of 'degree' and returns its root.

attrType here gives the type of he key
('i' for int,'f' for float and 'c' for string)
attLength gives the size of the key
(4 for int,4 for float and 1-255 for the string)

struct item{
                    'attrType' key;
                    struct node *p;
                    };

                    struct node{
                    struct node *p0;
                    struct item matrix[2*('degree')];
                    };

                    Where ' ' indicate a parameter to function
                    CreateTree(root,degree,attrType,attrLength)

Please help me... How can I do this????
I think this problem is impossible....

ps. (I think unions can't be used since I must pass the       key length...)

Thanks for your help...(if any)
kkarnezAsked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

kkarnezAuthor Commented:
Lets see your programming strength guys...

0
rbrCommented:
Try this 2 functions.

typedef struct node {
    void *pdata;  
    struct node **pplinks;
} NODE;


typedef struct {
    char type;
    int length;
    int degree;
    NODE *pnode;
} ROOT_NODE;


void CreateNodeEntry(ROOT_NODE *proot, NODE **ppnode)
{
    *ppnode = (NODE *)calloc (1, sizeof(NODE));
   
    (*ppnode)->pdata = calloc (proot->length, 1);
   
    (*ppnode)->pplinks=(struct node **)calloc (proot->degree, sizeof(NODE *));
}

void CreateTree (ROOT_NODE **pproot, int degree, char attrType, int attrLength)
{
    *pproot = (ROOT_NODE *)calloc (1, sizeof(ROOT_NODE));
   
    (*pproot)->type=attrType;
    (*pproot)->degree=degree;
    (*pproot)->length=attrLength;
    CreateNodeEntry (*pproot, &((*pproot)->pnode));
}


CreateTree will create your tree and with CreateNodeEntry you will allocate memory for every node you need. There are no error checkings inside these to functions (Enough memory free, attrLength == 4 for attrType == 'i', ...) but I hope this will help to solve your problems.

Robert B. Rossmann


0
LucHoltkampCommented:
I suggest you move to C++, then you can do this in a typesave and elegant way, using templates.

Furthermore I think rbr's idea of using pointers is the best way to go in C.
.luc.
0
Redefine Your Security with AI & Machine Learning

The implications of AI and machine learning in cyber security are massive and constantly growing, creating both efficiencies and new challenges across the board. Check out our on-demand webinar to learn more about how AI can help your organization!

y_lestrCommented:
I have all the sources to create and use Btrees in Pascal, I have done it when I was student, it's possible to easily translate the sources in C
0
jos010697Commented:
The problem can be solved, although in a 'dirty' way. What
you actually want, is an array of things embedded in a struct,
while the size of the array is not known in advance. For
the sake of the example, here's a simplified structure:

struct varlen_t {
   int size;
   <type> a[1];
}

We're assuming at least one element present in the array. Here's
a way to allocate a structure containing an array with 'n' elements:

varlen_t* struct_alloc(int n) {

varlen_t* v;

v== malloc(sizeof(struct varlen_t)+(n-1)*(sizeof(<type>));

v->size= n;

return v;

}

Note that I simply allocated n-1 elements more for the
entire thingy. This'll work for any array being the last
element of a struct ... (ANSI/ISO tried to rule out this
'feature', but they only succeeded half way ;-)

Also note that you have to store the actual size of the
array in the structure somewhere (nowhere else the
size is known, so you have to ...)

kind regards,

Jos aka jos@and.nl
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
stardateCommented:
I was looking at writing a b-tree myself, but opted for some source code that I found in simtel.  This code will do the job I need, with minor tinkering.  You may find that this code will do what you need.  The code will most certainly demonstrate how to create a (new)root node.

This source code is located at this URL

http://oak.oakland.edu/pub/simtelnet/msdos/c/bplus11.zip

Please note that this is shareware code.
0
nate091597Commented:
It's not impossible, but it sounds like an assignment.
0
NewGuyCommented:
Won't things become easier if we use

typedef ROOT_NODE *ROOT_NODEPTR;

and reduce all the pointer stuff
gosh...see less stars then
0
qed070297Commented:
C's typing implementation does not allow for this degree of
flexibility from mere declarations.  In particular, you cannot
pass types as parameters to functions.  All typing is considered
fixed in one way or another.  I would recommend a more
brute force mechanism for solving this problem:

struct keytag {
    int attrLength;
    char KeyContents[];
};

struct itemtag {
    struct node *p;
    struct keytag *key;
};

struct node {
    struct node *p0;
    int degrees;
    struct item matrix[];
};

struct node * CreateNode(int degree,int attrLength) {
int i;
struct node * tmp;

    tmp = (struct node *)malloc(sizeof(struct node)+\
                                degree*sizeof(struct item));
    for(i=0;i<degree;i++) {
        tmp->matrix[i].node = tmp;
        tmp->matrix[i].key = (struct keytag*)\
            malloc(sizeof(struct keytag)+\
                   attrLength*sizeof(char));
        tmp->matrix[i].key->attrLength = attrLength;
    }

    return tmp;
}

And so on.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
C

From novice to tech pro — start learning today.