Question

When I my function, my hash is empty. Why?

Here is my code:

#include <stdio.h>
#include <string.h>
#include "uthash.h"
struct oid_struct {
  char descr[20];
  char oid[50];
  UT_hash_handle hh;
};

testadd( struct oid_struct* oid_hash){

 struct oid_struct *element;
 element=(struct oid_struct*) malloc(sizeof(struct oid_struct));

 strcpy(element->descr, "foo");
 strcpy(element->oid, "1.2.1.34");
 HASH_ADD_STR(oid_hash, descr, element);
 printf("Hash has %d entries\n",HASH_COUNT(oid_hash));

}


main(){
        struct oid_struct *oid_hash = NULL, *lookup;
        testadd(oid_hash);
        printf("Hash has %d entries\n",HASH_COUNT(oid_hash));

}

Here is the output:

# gcc hashtest.c
# ./a.out
Hash has 1 entries
Hash has 0 entries
#

Answer 1

C passes arguments by value, meaning a copy of oid_hash is being changed inside testadd() so the change is invisible to the caller. Pass the address of oid_hash to testadd():

testadd(&oid_hash);

void testadd(struct oid_struct** oid_hash)
{
    *oid_hash = element; /* Depending on what is going on
                            inside HASH_ADD_STR(). */
}

Note the casting of return value of malloc() is not required.