Initializing C struct using pointer to pointer

Question

I'm working on creating a hash table implementation for an assignment. I've defined my hashtable as struct as follows:

typedef struct hashtable { int size; int entries; int table*; // pointer to table. Each entry will point to linked list // of key-value nodes } hashtable;

I have to initialize the hashtable struct in a method using double pointers, e.g.:

void init(hashtable** ht) { ... }

I've written a basic implementation below:

#include <stdio.h> #include <stdlib.h> typedef struct hashtable { int size; int entries; int table*; // pointer to table. Each entry will point to linked list // of key-value nodes } hashtable; void init(hashtable**); void init(hashtable** ht) { *ht = (hashtable *) malloc( sizeof(hashtable) ); *ht->size = 3; } int main(){ hashtable *t = NULL; init(&t); printf("t.size: %i", t->size); }

However, I keep getting the following compile error:

ll.c:19:8: error: member reference base type 'hashtable *' (aka 'struct hashtable *') is not a structure or union *ht->size = 3; ~~^ ~~~~ 1 error generated.

So I'm confused by the following: 1. I'm not sure how to create a new struct in the init function when being passed a pointer to pointer. 2. After allocating the struct, how do I modify the struct member attributes?

this *ht->size = 3; --> (*ht)->size = 3;

Seek Addo
– Seek Addo

2017-07-12 22:46:17 +00:00
Commented Jul 12, 2017 at 22:46 — Seek Addo
– Seek Addo, Commented Jul 12, 2017 at 22:46
int table*;???

AnT stands with Russia
– AnT stands with Russia

2017-07-12 22:54:11 +00:00
Commented Jul 12, 2017 at 22:54 — AnT stands with Russia
– AnT stands with Russia, Commented Jul 12, 2017 at 22:54

PraMiD · Accepted Answer · 2017-07-12 22:53:10Z

This is just a operator precedence problem.

The compiler processes -> before the *. Therefore, it tries to access the size member of struct hashtable ** which is not possible.

The code compiles if you exchange *ht->size with (*ht)->size.

Zakir · Accepted Answer · 2017-07-12 22:51:15Z

You have 2 errors in your code :

int table* --> int *table - Declare pointer to integer
*ht->size --> (*ht)->size - Imp to put brackets when you are not sure of operator precedence

Alexander · Accepted Answer · 2017-07-12 22:55:10Z

That's a good start, and others have addresses the primary issues in your code. However, I would suggest a minor tweak:

#include <stdio.h> #include <stdlib.h> typedef struct hashtable { int size; int entries; int table*; // pointer to table. Each entry will point to linked list // of key-value nodes } hashtable; // note: freeing the hashtable is a caller responsibility! hashtable *new_hashtable() { hashtable *ht = malloc( sizeof(hashtable) ); ht->size = 3; // since ht is only a single pointer, no need for (*ht)->foo return ht; } int main(){ hashtable *ht = new_hashtable(); printf("ht.size: %i", ht->size); free(ht); }

appreciate the feedback on returning the pointer. I'm constrained here by my assignment specifications.
Oh I see. Well just keep this in mind for the future. This way is safer, because you're never meant to assign NULL. You just get the pointer directly from the function, without a chance to forget to call init.

Davide Spataro · Accepted Answer · 2017-07-12 23:01:16Z

The problem is that

-> has higher precedence than * in C as you can see from here

using precedence rules *ht->size translates to *(ht->size). That should make clear the reason why you get the error. Another way to see it is

*(ht->size)=(*(*ht).size)

Fix this using parenthesis as follows: (*ht)->size

There is another issue in the definition of hashtable: int table*; won't compile. Use int *table; instead to declare a pointer to int?

Dhairya Dalal · Accepted Answer · 2017-07-12 23:24:32Z

Thanks all for the quick response. For future reference, here's a quick update of the original code with solutions:

#include <stdio.h> #include <stdlib.h> typedef struct hashtable { int size; // size of hash table int entries; // number of slots allocated in table int *table; /* pointer to table. Each entry will point to linked list of key-value nodes */ } hashtable; void init(hashtable**); void init(hashtable** ht) { *ht = (hashtable *) malloc( sizeof(hashtable) ); (*ht)->entries = 0; (*ht)->size = 3; //replace this with better init size, ideally a prime number (*ht)->table = malloc( (*ht)->size * sizeof(int)); } int main(){ hashtable *t = NULL; init(&t); t->table[2] = 3; printf("t.size: %i \n", t->size); printf("t.arr:[2] %i \n", t->table[2]); free(t); }

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