C - 链表打印顺序错误



我创建了一个链表,其元素是从命令行参数获取的字符串:

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    struct element_Args {
        char commandLineArgs[500];
    };

    struct list {
    struct element_Args element;
    struct list *next;
    };
    int main(int argc, char *argv[]) {
    struct list *head;
    struct list *current;
    head = (struct list *) malloc(sizeof(struct list));

    head->next = NULL;
    int i;
    for(i = 0; i < argc; i++) {
        current = malloc (sizeof(struct list));
        strcpy(current->element.commandLineArgs, argv[i]);
        current->next = head;
        head = current;
    }
    current = head;
    while(current->next != NULL) {
       printf("%sn", current->element.commandLineArgs);
       current = current->next;
    }
    return 0;
    }

但是,当我打印链表中的元素时,它们以与输入参数的顺序相反的方式打印出来。 如何按照输入时的顺序打印它们? 我觉得我好像错过了一些小东西,但我无法弄清楚那是什么。

在 for 循环中,删除head = current

基本上,使用这条线会忘记你的头。稍后可以通过设置临时指针遍历head,但不要重置head(除非插入新head)。

要插入一个新的头部,你会说,newHead->next = head; head = newHead;如果你想按顺序插入它们,你应该保留一个尾部指针,并始终在末尾添加。

int i;
struct list* tail = head;
for(i = 0; i < argc; i++) {
    current = malloc (sizeof(struct list));
    if(current != NULL){
        strcpy(current->element.commandLineArgs, argv[i]);
        tail->next =  current; // add this line
        tail = tail->next;
        current->next = head; //this line makes you add in reverse order. Remove this as well.
        head = current; // remove this line here
    }

}
这是

你的问题

head = current;

您应该head指向第一个节点,并且永远不要再次覆盖它。

因此,在您的代码中,head实际上是tail,即以相反的顺序打印值是合乎逻辑的,您期望相反的情况并非如此。

试试这个

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct element_Args {
    char commandLineArgs[500];
};

struct list {
struct element_Args element;
struct list *next;
};
int main(int argc, char *argv[]) {
    struct list *head;
    struct list *current;
    struct list *last;
    int i;
    head = malloc(sizeof(*head));
    if (head == NULL)
        return -1;
    head->next = NULL;
    last = head;
    for(i = 0 ; i < argc ; i++) {
        current = malloc (sizeof(*current));
        if (current != NULL) {
            strcpy(current->element.commandLineArgs, argv[i]);
            last->next = current;
            last       = current;
        }
    }
    current = head;
    while(current != NULL) {
        printf("%sn", current->element.commandLineArgs);
        current = current->next;
    }
    return 0;
}

不要忘记编写一个freeList()函数来free所有malloc编辑的内容。

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