如何修改傳遞給C中函數的指針？因此，我有一些代碼，類似于下面的代碼，可以將結構添加到結構列表中：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}這些結構的定義如下：typedef struct Bar{
// this isn't too important} Bar;#define EMPTY_LIST NULLtypedef struct BarList{
Bar * val;
struct BarList * nextBar;} BarList;然后在另一個文件中執行如下操作：BarList * l;l = EMPTY_LIST;barPush(l,&b1); // b1 and b2 are just Bar'sbarPush(l,&b2);但是，在此之后，l仍然指向空_list，而不是barPush內部創建的修改版本。如果我想修改一個指針，是否必須將列表作為指針傳入，還是需要使用其他暗咒語？
3 回答
函數式編程
TA貢獻1807條經驗 獲得超9個贊
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; }BarList * l;l = EMPTY_LIST;barPush(&l,&b1); // b1 and b2 are just Bar'sbarPush(&l,&b2);
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; }BarList * l;l = EMPTY_LIST;l = barPush(l,&b1); // b1 and b2 are just Bar'sl = barPush(l,&b2);
幕布斯6054654
TA貢獻1876條經驗 獲得超7個贊
int myFunction(int** param1, int** param2) {// now I can change the ACTUAL pointer - kind of like passing a pointer by reference }- 3 回答
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