【发布时间】:2016-08-31 10:52:00
【问题描述】:
我正在尝试制作一个按字母顺序排序的链表。对于代码,如果任何两个名称相同,则比较它们的年龄。假设他们的年龄不会相同。
我创建了一个函数来比较两个学生的姓名和年龄,如 comeBefore 所示。这很好用。
我遇到的问题是将四个名称链接在一起时出现分段错误。如果有人能给我一些关于我哪里出错或在 c 中调试的指示,那就太好了。
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <ctype.h>
typedef struct name_s Name;
//Struct name_s contains a character pointer to a name, an age, and a
// pointer to the next name in a linked list
struct name_s{
char* name;
int age;
Name* next;
};
//comesBefore takes a pointer to student1 and student2.
//The two students are compared to see which one comes first in
//alphabetical order.
//If both names are the same then their ages are compared.
//The function returns true if student1 should come first else false
bool comesBefore(const Name* student1, const Name* student2)
{
int i = 0;
if (student2 == NULL){
return true;
}
while (student1->name[i]!= '\0' && student2->name[i] != '\0'){
if (student1->name[i] < student2->name[i]){
return true;
}
i++;
}
if (student1->name == student2->name){
if (student1->age < student2->age){
return true;
}
}
return false;
}
//creates a linked list in alphabetical order of the names
//if any two names are the same the one with the lowest age comes first
Name* linkedList(Name names[], int n)
{
Name* head = NULL;
Name* previous = NULL;
Name* last = NULL;
Name* current = &names[0];
int i = 0;
while (i < n) {
if (head == NULL){
head = current;
} else {
if (comesBefore(current, head)){
current->next = head;
head = current;
} else {
previous = head;
while(comesBefore(current, previous->next) == false
&& previous->next != NULL){
previous = previous->next;
}
current->next = previous->next;
previous->next = current;
}
}
last = head;
int k = 0;
while (k <= i){
last = last->next;
k++;
}
last->next = NULL;
i++;
current = &names[i];
}
return head;
}
int main(void)
{
Name a;
Name b;
Name c;
Name d;
Name* s1 = &a;
Name* s2 = &b;
Name* s3 = &c;
Name* s4 = &d;
s1->name = "Zaphod Beeblebrox";
s1->age = 250;
s2->name = "Albert Einstein";
s2->age = 133;
s3->name = "Albert Einstein";
s3->age = 7;
s4->name = "Brook Fleming";
s4->age = 20;
Name names[4];
names[0] = a;
names[1] = b;
names[2] = c;
names[3] = d;
Name* list = linkedList(names, 4);
while (list!=NULL){
printf("Name: %s\nAge: %d\n--------\n",
list->name,
list->age
);
list = list->next;
}
return 0;
}
【问题讨论】:
-
你的比较功能有问题。假设代码将 Zaphod 与 Albert 进行比较;第一次比较发现 Z 不小于 A,因此继续比较
a和l— 结果错误。您需要测试if (student1->name[i] > student2->name[i]) return false;,并且仅在到目前为止名称相等时才继续循环。随后的if (student1->name == student2->name)测试也是假的;比较指针是否相等不是你想要的——你必须再次测试小于和大于,然后决定如果年龄相等则返回什么。
标签: c data-structures struct linked-list alphabetical