Question

I made this little piece of code where I alloc and then free memory for a struct:

struct item {
    int val;
    int *vectOfInt;
    struct item *next;
};

void relItem(struct item **currItem) {
    struct item *temp;
    int *intTemp;

    temp = *currItem;
    intTemp = (*currItem)->vectOfInt;
    *currItem = (*currItem)->next;
    free(temp);
    free(intTemp);
}

int main() {
    int array[] = {0, 1, 2, 3, 4, 5};
    struct item *list = NULL;

    list = (struct item*) malloc(sizeof(struct item));
    list->val = 0;
    list->vectOfInt = array;
    list->next = NULL;

    relItem(&list);

    return 0;
}

EDIT: code for the comment:

struct item {
    int val;
    struct item *next;
};

void edit(struct item *currItem) {
    currItem->val = 2;
}

int main() {
    struct item *list = NULL;
    list = (struct item*) malloc(sizeof(struct item));
    list->val = 0;
    list->next = NULL;

    edit(list);

    //list-val == 2

    return 0;
}

• How can I do the same thing without using double-pointer to the struct?

• Can you explain me why and how this works (both with pointer and double-pointer)?

• I don’t understand how a struct is represented in main memory
  (for example int a[5]; in memory a is a pointer to the first location of the buffer allocated for the a[5] array)

what is the equal representation for a struct initialized wit ha pointer
(struct item *s) ?

Answer 1

-You need to use double pointer since you need to change the pointer to your struct and call by reference is not allowed in C.

-It would not work with a single pointer. Because you want to change the pointer to your object. You do it by the statement *currItem = (*currItem)->next;
In order to change it permanently, you need to use a pointer to it. Which makes you use double pointer.

Think of it that way:

you have an integer variable a that you want a function to change its value. You simply call this function with a pointer to the variable a
like:

void changeTheValue(int *x)
{
    *x = 7;
}
void main()
{
   int a = 5;
   changeTheValue(&a);
}

and in your case you want to change your pointer’s value and you simply pass its pointer to the function.(double pointer)
simple as that. If you want to change value of something with a function, then you have to pass its pointer to the function.

-When you make a call to malloc, you want space from it. And you state that you want a space as big as the size of your struct. (Like when you did in here list = (struct item*) malloc(sizeof(struct item));) and malloc allocates a space as big as your struct. If size of your struct is 1 byte then you have a 1 byte space, if it is 4, then you have consecutive 4 bytes.
See, this is how your struct is kept in the memory.
If you declare a struct variable or an array etc. then your struct is kept in memory like (dunno if it is a good metaphor) an array. The first member comes first, and then the second…

say if you have a struct

struct myStruct
{
    int a;
    float b;
    char c;
};

then the memory looks like

---

a

---

b

---

c

ps: you are making an illegal call to free on the line free(intTemp); you are *free*ing a variable that you didn’t *malloc*ed.