如何修改已传递给 C 中函数的指针? [英] How do I modify a pointer that has been passed into a function in C?
问题描述
所以,我有一些类似于以下的代码,用于将结构添加到结构列表中:
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
}
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如下操作:
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);
但是,在这之后,l 仍然指向 EMPTY_LIST,而不是在 barPush 内部创建的修改版本.如果我想修改它,我是否必须将列表作为指向指针的指针传入,或者是否需要其他一些黑暗的咒语?
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?
推荐答案
如果你想这样做,你需要传入一个指向指针的指针.
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;
}
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:
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;
}
用法变成:
BarList * l;
l = EMPTY_LIST;
l = barPush(l,&b1); // b1 and b2 are just Bar's
l = barPush(l,&b2);
这篇关于如何修改已传递给 C 中函数的指针?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!