How do I modify a pointer that has been passed into a function in C?

Question

So, I have some code, kind of like the following, to add a struct to a list of structs:

void barPush(BarList * list,Bar * bar) {     // if there is no move to add, then we are done     if (bar == NULL) return;//EMPTY_LIST;      // allocate space for the new node     BarList * newNode = malloc(sizeof(BarList));      // assign the right values     newNode->val = bar;     newNode->nextBar = list;      // and set list to be equal to the new head of the list     list = newNode; // This line works, but list only changes inside of this function } 

These structures are defined as follows:

typedef struct Bar {     // this isn't too important } Bar;  #define EMPTY_LIST NULL  typedef struct BarList {     Bar * val;     struct  BarList * nextBar; } BarList; 

and then in another file I do something like the following:

BarList * l;  l = EMPTY_LIST; barPush(l,&b1); // b1 and b2 are just Bar's barPush(l,&b2); 

However, after this, l still points to EMPTY_LIST, not the modified version created inside of barPush. Do I have to pass list in as a pointer to a pointer if I want to modify it, or is there some other dark incantation required?

Answer 1

You need to pass in a pointer to a pointer if you want to do this.

void barPush(BarList ** list,Bar * bar) {     if (list == NULL) return; // need to pass in the pointer to your pointer to your list.      // if there is no move to add, then we are done     if (bar == NULL) return;      // allocate space for the new node     BarList * newNode = malloc(sizeof(BarList));      // assign the right values     newNode->val = bar;     newNode->nextBar = *list;      // and set the contents of the pointer to the pointer to the head of the list      // (ie: the pointer the the head of the list) to the new node.     *list = newNode;  } 

Then use it like this:

BarList * l;  l = EMPTY_LIST; barPush(&l,&b1); // b1 and b2 are just Bar's barPush(&l,&b2); 

Jonathan Leffler suggested returning the new head of the list in the comments:

BarList *barPush(BarList *list,Bar *bar) {     // if there is no move to add, then we are done - return unmodified list.     if (bar == NULL) return list;        // allocate space for the new node     BarList * newNode = malloc(sizeof(BarList));      // assign the right values     newNode->val = bar;     newNode->nextBar = list;      // return the new head of the list.     return newNode;  } 

Usage becomes:

BarList * l;  l = EMPTY_LIST; l = barPush(l,&b1); // b1 and b2 are just Bar's l = barPush(l,&b2);