How do I modify a pointer that has been passed into a function in C?
Solution 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);
Solution 2
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.
Solution 3
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
}
Solution 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.
Solution 5
Yes, you have to pass in a pointer to the pointer. C passes arguments by value, not by reference.
Paul Wicks
Updated on July 10, 2022Comments
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Paul Wicks almost 2 years
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?
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Paul Wicks about 15 yearsThanks, I'd figured this was the problem, but hoped that it wasn't ;)
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Jonathan Leffler about 15 yearsAlternatively, have the function return the pointer to the new head of the list. BarList *barPush(BarList *list, Bar *bar)