C ++声明双链表,用两个for循环遍历双向列表并打印
C++ declaring double-linked list, traverse list both ways with two for-loops and print
刚开始接触 C++(请温柔点)。我有一些代码,我必须:
- 添加元素* 构成双向链表
- 向元素添加 insert_before() 和 insert_after() 方法
- 使用两个单独的 for 循环遍历列表的两种方式
- 在每个循环中打印操作数
代码
我尝试了以下方法:
#include <cstdio>
struct Element {
Element* next{};
Element* previous{};
void insert_after(Element* new_element) {
new_element -> previous = this;
new_element -> next = this -> next;
this -> next = new_element;
}
void insert_before(Element* new_element) {
new_element -> previous = this -> previous;
new_element -> next = this;
this -> previous = new_element;
}
char prefix[2];
short operating_number;
};
int main() {
Element trooper1, trooper2, trooper3, trooper4;
trooper1.prefix[0] = 'T';
trooper1.prefix[1] = 'K';
trooper1.operating_number = 421;
trooper1.insert_after(&trooper2);
trooper2.prefix[0] = 'F';
trooper2.prefix[1] = 'N';
trooper2.operating_number = 2187;
trooper2.insert_before(&trooper3);
trooper3.prefix[0] = 'L';
trooper3.prefix[1] = 'S';
trooper3.operating_number = 005;
trooper3.insert_before(&trooper4);
trooper4.prefix[0] = 'F';
trooper4.prefix[1] = 'K';
trooper4.operating_number = 2602;
for (Element *cursor = &trooper1; cursor; cursor = cursor -> next) {
printf("stormtrooper %c%c-%d\n",
cursor->prefix[0],
cursor->prefix[1],
cursor->operating_number);
}
for (Element *cursor = &trooper3; cursor; cursor = cursor -> previous) {
printf("stormtrooper %c%c-%d\n",
cursor->prefix[0],
cursor->prefix[1],
cursor->operating_number);
}
}
这是非常基础的,我知道。但我开始学习了。
代码编译正确但我得到这个输出:
/home/facundo/Escritorio/C++Projects/cmake-build-debug/C__Projects
stormtrooper TK-421
stormtrooper FN-2187
stormtrooper LS-5
stormtrooper FK-2602
stormtrooper TK-421
Process finished with exit code 0
我真的不明白为什么它只打印那个(我猜应该有8行输出)。
我期待的输出会打印这样的元素:
士兵 1
骑兵4
骑兵3
士兵2
士兵2
骑兵3
骑兵4
士兵 1
如果能提供一些帮助,我们将不胜感激。感谢您抽出时间。
在您的 insert
函数中,您没有正确链接所有指针:
void insert_after(Element* new_element) {
new_element -> previous = this;
new_element -> next = this -> next;
this -> next = new_element;
if (new_element -> next) // this check needed
// to correctly link the next -> previous
new_element -> next -> previous = new_element;
}
和
void insert_before(Element* new_element) {
new_element -> previous = this -> previous;
new_element -> next = this;
this -> previous = new_element;
if (new_element -> previous) // this check needed
// to correctly link the previous -> next
new_element -> previous -> next = new_element;
}
此外,在第二个 for
循环中,请注意 trooper2
是列表中的最后一个 Element
,而不是 trooper3
,因此您需要从那里开始以反向查看所有 Element
。
这里是 demo。
刚开始接触 C++(请温柔点)。我有一些代码,我必须:
- 添加元素* 构成双向链表
- 向元素添加 insert_before() 和 insert_after() 方法
- 使用两个单独的 for 循环遍历列表的两种方式
- 在每个循环中打印操作数
代码
我尝试了以下方法:
#include <cstdio>
struct Element {
Element* next{};
Element* previous{};
void insert_after(Element* new_element) {
new_element -> previous = this;
new_element -> next = this -> next;
this -> next = new_element;
}
void insert_before(Element* new_element) {
new_element -> previous = this -> previous;
new_element -> next = this;
this -> previous = new_element;
}
char prefix[2];
short operating_number;
};
int main() {
Element trooper1, trooper2, trooper3, trooper4;
trooper1.prefix[0] = 'T';
trooper1.prefix[1] = 'K';
trooper1.operating_number = 421;
trooper1.insert_after(&trooper2);
trooper2.prefix[0] = 'F';
trooper2.prefix[1] = 'N';
trooper2.operating_number = 2187;
trooper2.insert_before(&trooper3);
trooper3.prefix[0] = 'L';
trooper3.prefix[1] = 'S';
trooper3.operating_number = 005;
trooper3.insert_before(&trooper4);
trooper4.prefix[0] = 'F';
trooper4.prefix[1] = 'K';
trooper4.operating_number = 2602;
for (Element *cursor = &trooper1; cursor; cursor = cursor -> next) {
printf("stormtrooper %c%c-%d\n",
cursor->prefix[0],
cursor->prefix[1],
cursor->operating_number);
}
for (Element *cursor = &trooper3; cursor; cursor = cursor -> previous) {
printf("stormtrooper %c%c-%d\n",
cursor->prefix[0],
cursor->prefix[1],
cursor->operating_number);
}
}
这是非常基础的,我知道。但我开始学习了。 代码编译正确但我得到这个输出:
/home/facundo/Escritorio/C++Projects/cmake-build-debug/C__Projects
stormtrooper TK-421
stormtrooper FN-2187
stormtrooper LS-5
stormtrooper FK-2602
stormtrooper TK-421
Process finished with exit code 0
我真的不明白为什么它只打印那个(我猜应该有8行输出)。 我期待的输出会打印这样的元素:
士兵 1
骑兵4
骑兵3
士兵2
士兵2
骑兵3
骑兵4
士兵 1
如果能提供一些帮助,我们将不胜感激。感谢您抽出时间。
在您的 insert
函数中,您没有正确链接所有指针:
void insert_after(Element* new_element) {
new_element -> previous = this;
new_element -> next = this -> next;
this -> next = new_element;
if (new_element -> next) // this check needed
// to correctly link the next -> previous
new_element -> next -> previous = new_element;
}
和
void insert_before(Element* new_element) {
new_element -> previous = this -> previous;
new_element -> next = this;
this -> previous = new_element;
if (new_element -> previous) // this check needed
// to correctly link the previous -> next
new_element -> previous -> next = new_element;
}
此外,在第二个 for
循环中,请注意 trooper2
是列表中的最后一个 Element
,而不是 trooper3
,因此您需要从那里开始以反向查看所有 Element
。
这里是 demo。