如何修改已传递到C函数中的指针？

Question

所以,我有一些代码,类似于以下,将结构添加到结构列表:

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
}

这些结构定义如下:

typedef struct Bar
{
    // this isn't too important
} Bar;

#define EMPTY_LIST NULL

typedef struct BarList
{
    Bar * val;
    struct  BarList * nextBar;
} BarList;

然后在另一个文件中,我执行以下操作:

BarList * l;

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

但是,在此之后,l仍然指向EMPTY_LIST,而不是barPush内部创建的修改版本.如果我想修改它,或者是否需要其他暗咒语,我是否必须将列表作为指针传递给指针？

Answer 1

如果要执行此操作,则需要传入指针指针.

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; 
}

然后像这样使用它:

BarList * l;

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

Jonathan Leffler建议在评论中返回名单的新负责人:

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; 
}

用法变为:

BarList * l;

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

Answer 2

通用答案:将指针传递给您想要更改的内容.

在这种情况下,它将是指向您想要更改的指针的指针.

Answer 3

请记住,在C中,一切都按值传递.

你传入指向指针的指针,就像这样

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

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

}

Answer 4

这是一个经典问题。要么返回分配的节点，要么使用指针的指针。在 C 中，您应该将指向 X 的指针传递给您想要修改 X 的函数。在这种情况下，由于您想要修改指针，因此应该将指针传递给指针。