C++ pointers doubts

Question

I have a doubt about this code I saw at the university.

struct nodeList{
  int data;
  nodeList * next;
};
typedef nodeList * List;

void filter( List &l )
{
   List * aux = &l;
   while(*aux)
   {
       if( (*aux)->data > 3 )
       {
          List toDelete = *aux;
          *aux = (*aux)->next;
          delete toDelete;
       }
       else
       {
          aux = &((*aux)->next);
       }   
   }
}

• I don't know, what does List * aux = &l; actually do. Because List is the same as nodeList * so, the code would be nodeList * * aux = &l; and that is actually what I don't understand, is that a pointer to a pointer that holds the address of a pointer of a nodeList struct?

• The second thing I have trouble understanding is the last line. aux = &((*aux)->next); Why is aux without a * on the left side? if It was declared as List *aux = &l; is List just a pointer to the first node?

Thank you in advance. I googled a lot, but I didn't find any answers. If you can answer my questions I'll appreciate a lot.

Answer 1

1. Exactly.

2. You should always match the datatypes. aux is a nodeList **, therefore any assignment should be of the same datatype. Since ((*aux)->next) is nodeList *, you use the & operator to get a nodeList **.

Datatypes

Variables have specific datatypes. For example, aux is a List* but List is an alias of nodeList*, so aux is a nodeList**.

But also expressions have datatypes as a whole. For example,

• the expression ((*aux)->next) is of datatype nodeList * and &((*aux)->next) is of datatype nodeList **. You use the operator & to get the memory address of a variable, and the result data type of using & is one more star.
• the expression *aux is of datatype nodeList *, because aux is nodeList ** and the star operator gets the value of the pointed element by the pointer, effectively removing one star from the datatype.