[英]Pointers to pointers: partitioning a linked list
我正在尝试变得更容易使用指针。 因此,为了好玩,我采用了以下C ++函数,该函数将链接列表划分为一个值
void partitionList(lnode<int> *& head, int val) {
lnode<int> * front = nullptr;
lnode<int> * back = nullptr;
lnode<int> * curr = head;
while (curr) {
lnode<int> * next = curr->next;
if (curr->data < val) {
curr->next = front;
front = curr;
} else {
curr->next = back;
back = curr;
}
curr = next;
}
curr = front;
while (curr->next) {
curr = curr->next;
}
curr->next = back;
head = front;
}
我试图将其更改为采用C样式的双指针。 我做了一个漫不经心的查找替换,但是没有用。 经过调查,我找到了问题的根源,但我仍然不太了解发生了什么...
void partitionList(lnode<int> ** head, int val) {
lnode<int> * front = nullptr;
lnode<int> * back = nullptr;
lnode<int> ** curr = head;
while (*curr) {
lnode<int> * entry = *curr;
std::cout << (*curr)->data << std::endl; // On second loop, prints 2
std::cout << entry->data << std::endl; // On second loop, prints 2
lnode<int> * next = entry->next; // This assignment does something
std::cout << entry->data << std::endl; // On second loop, prints 2
std::cout << (*curr)->data << std::endl; // On second loop, prints 3!
if ((*curr)->data < val) {
(*curr)->next = front;
front = *curr;
} else {
(*curr)->next = back;
back = *curr;
}
curr = &next;
}
*curr = front;
while ((*curr)->next) {
(*curr) = (*curr)->next;
}
(*curr)->next = back;
head = &front;
}
int main() {
lnode<int> * tail = new lnode<int>(8, nullptr);
lnode<int> * seven = new lnode<int>(7, tail);
lnode<int> * six = new lnode<int>(6, seven);
lnode<int> * five = new lnode<int>(5, six);
lnode<int> * four = new lnode<int>(4, five);
lnode<int> * three = new lnode<int>(3, four);
lnode<int> * two = new lnode<int>(2, three);
lnode<int> * head = new lnode<int>(1, two);
partitionList(&head, 6);
}
在第一个循环中,在函数while循环顶部附近的所有四个调试打印行均打印“ 1”。 但是,在第二个循环中,它们显示“ 2”,“ 2”,“ 2”,“ 3”吗?
谁能解释这是怎么回事? 使用双指针而不是对指针的引用的正确方法是什么?
谢谢!
void partitionList(lnode<int> *& head, int val) {
...
head = front; // <= head is modified
}
但是这里不是
void partitionList(lnode<int> ** head, int val) {
...
head = &front; // try it with *head = front
}
如果您想对此进行快速替换
void partitionList(lnode<int> *& head, int val) {
...
lnode<int> * curr = head;
只需这样写:
void partitionList(lnode<int> ** head, int val) {
...
lnode<int> * curr = *head;
为什么将*cur
修改为**cur
?
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