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

Yes, you have to pass in a pointer to the pointer. C passes arguments by value, not by reference.

Answer 2

Remember, in C, EVERYTHING is passed by value.

You pass in a pointer to a pointer, like this

int myFunction(int** param1, int** param2) {

// now I can change the ACTUAL pointer - kind of like passing a pointer by reference 

}

Answer 3

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);

Answer 4

This is a classic problem. Either return the allocated node or use a pointer of pointer. In C, you should pass a pointer to a X to a function where you want your X to be modified. In this case, since you want a pointer to be modified, you ought to pass a pointer to a pointer.

Answer 5

Generic answer: Pass a pointer to the thing you want to change.

In this case, it would be a pointer to the pointer you want to change.