Passing struct pointer to function in c

Question

I'm having a problem with passing a pointer to a struct to a function. My code is essentially what is shown below. After calling modify_item in the main function, stuff == NULL. I want stuff to be a pointer to an item struct with element equal to 5. What am I doing wrong?

void modify_item(struct item *s){
   struct item *retVal = malloc(sizeof(struct item));
   retVal->element = 5;
   s = retVal;
}

int main(){
   struct item *stuff = NULL;
   modify_item(stuff); //After this call, stuff == NULL, why?
}

Answer 1

Because you are passing the pointer by value . The function operates on a copy of the pointer, and never modifies the original.

Either pass a pointer to the pointer (ie a struct item ** ), or instead have the function return the pointer.

Answer 2

void modify_item(struct item **s){
   struct item *retVal = malloc(sizeof(struct item));
   retVal->element = 5;
   *s = retVal;
}

int main(){
   struct item *stuff = NULL;
   modify_item(&stuff);

or

struct item *modify_item(void){
   struct item *retVal = malloc(sizeof(struct item));
   retVal->element = 5;
   return retVal;
}

int main(){
   struct item *stuff = NULL;
   stuff = modify_item();
}

Answer 3

I would suggest to modify your code like below if the function 'modify_item' intends to change member of structure which is passed as argument.

void modify_item(struct item *s){
   s->element = 5;
}

int main(){
   struct item *stuff = malloc(sizeof(struct item));
   modify_item(stuff); 
}

Answer 4

You can also make a reference to pointer:

void modify_item(struct item* &s){
   struct item *retVal = (struct item*)malloc(sizeof(struct item));
   retVal->element = 5;
   s = retVal;
}

int main(){
   struct item *stuff = NULL;
   modify_item(stuff); //After this call, stuff == NULL, why?
    cout<<stuff->element;
    return 0;
}