c - 如何修改已传递到C函数中的指针？

因此，我有一些类似于以下内容的代码，可将结构添加到结构列表中：

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
}

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，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，此后，我仍然指向EMPTY_LIST，而不是在barPush内部创建的修改版本。如果要修改列表，是否必须将列表作为指针传递给指针，还是需要其他暗咒语？

最佳答案

如果要执行此操作，则需要将指针传递给指针。

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建议在评论中返回列表的新头：

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);