Initialize new Hash Table, pointers issue

Question

I have this exercise for homework that follows:

Consider the following definition to represent dynamic hashtables with collision treatment chaining.

typedef struct entry{
   char key[10];
   void *info;
   struct entry *next;
}   *Entry;

typedef struct hashT{
   int hashsize;
   Entry *table;
}   *HashTable;

Define `HashTable newTable(int hashsize) note that the memmory necessary must be allocated and that all table entrys must be initialized with empty list.

My proposition is this:

HashTable newTable(int hashSize){
   Entry *table = malloc(sizeof((struct entry)*hashSize));
   HashTable *h = malloc(sizeof(struct hashT));
   h->hashSize = hashSize;
   h->table = table;
   return h;
}

im pretty sure the logic is correct. my problem is with pointers. for example, sometime I see (char *), or in this case (table *) before the malloc function... Is this necessary?

and for the return, should I return h, or *h? and whats the difference?

thank you

Try compiling with warnings enabled. You should be seeing errors. Hiding pointers in typedefs is usually bad for the exact reasons seen here. For example `Entry *table` is a pointer to pointer to struct entry, yet you try to allocate space for `sizeof (struct entry) * hashSize`, though that too is misparenthesized. — Ilja Everilä, Dec 30 '17 at 15:17
This is C basics. What does you C book say about pointers and structures? What is not clear about its explanation? Did you check another textbook/resource? Ask you teacher? And never ever `typedef` a pointer to data types, it is worst practice and causes bad style! — too honest for this site, Dec 30 '17 at 16:05

score 0 · Accepted Answer · edited Jun 20 '20 at 09:12

Firstly,

Entry *table = malloc(sizeof((struct entry)*hashSize));

will be

Entry table = malloc(sizeof(struct entry)*hashSize);
                           ^^^    
                           Look at the parentheses here.

Also you will do the same change over here,

HashTable h = malloc(sizeof(struct hashT));

Same goes with HashTable. You forgot that you have hidden the pointers inside typedef which you shouldn't.

With the above mentioned changes the code will be

HashTable newTable(int hashSize){
   Entry table = malloc(sizeof(struct entry)*hashSize);
   HashTable h = malloc(sizeof(struct hashT));
   h->hashSize = hashSize;
   h->table = table;
   return h;
}

And Dont hide pointer behind typedef.(Wish I could write this in red color but yes this will save you from many many problems).

What is the type of `Entry*`?

It is of type struct entry **. Here in your case you don't need it while allocating. It will be an overkill if you do so.

What is the difference between `h` and `*h`?

Well typewise h is of type struct hashT** and so *h will be of type struct hashT*.

What shall you do next?

Compile all your code with warnings enabled. gcc -Wall -Werror progname.c
Check the return value of malloc.
Free dynamically allocated memory when done working with it.
Read How to debug small programs.
Grab a book.

Initialize new Hash Table, pointers issue

1 Answers1

What is the type of Entry*?

What is the difference between h and *h?

What shall you do next?

What is the type of `Entry*`?

What is the difference between `h` and `*h`?