c - C中的字符串链表

Question

我是 C 中链表的新手，我遇到的问题是我正在尝试创建一个字符串的链表，但是当我尝试打印该列表时，它会从两个不同的字符串中打印第一个字符。我想我弄乱了一些指针。请问有什么帮助吗？这是我的代码...

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


int main()
{
    typedef struct _song{char *songTitle; char *songAuthor; char *songNote; struct _song *next;}SONG;

    int songCount =4;


    char SongTitle[songCount];
    char AuthorName[songCount];
    char SongNotes[songCount];


    char songTitle0[21] = "19 problems";
    char songArtist0[21]="JayZ";
    char songNotes0[81]="JiggaWhoJiggaWhat";
    SongTitle[0]=*songTitle0;//points at string songTitle0
    AuthorName[0]=*songArtist0;
    SongNotes[0]=*songNotes0;

    char songTitle1[21] = "Cig Poppa";
    char songArtist1[21]="Biggie Smalls";
    char songNotes1[81]="I Luv it When you call me big poppa";
    SongTitle[1]=*songTitle1;
    AuthorName[1]=*songArtist1;
    SongNotes[1]=*songNotes1;

    SONG *CurrentSong, *header, *tail;

    int tempCount=0;
    header = NULL;

    for(tempCount=0;tempCount<songCount;tempCount++)
    {

        CurrentSong = malloc(sizeof(struct _song));
        CurrentSong->songTitle= &SongTitle[tempCount];
        CurrentSong->songAuthor=&AuthorName[tempCount];
        CurrentSong->songNote=&SongNotes[tempCount];

        if(header == NULL)
        {
            header=CurrentSong;//head points to first thing in memory
        }
        else
        {
            tail->next=CurrentSong;
        }
        tail = CurrentSong;//always the last thing in the list 
        tail->next=NULL;//the next pointer is null always

    }
    tempCount =0;
    for(CurrentSong=header; CurrentSong!=NULL; CurrentSong=CurrentSong->next)
                    {
                        printf("\n%d: ", tempCount);
                            printf("Title: %s ",CurrentSong->songTitle);


                        printf("Author: %s ",CurrentSong->songAuthor);
                        tempCount++;
                    }


    return 0;
}

Answer 1

SongTitle[0]=*songTitle0;//points at string songTitle0

评论不属实。您正在将第一个字符复制songTitle0到SongTitle.

你的设置太复杂了。您只需将songTitle0不带任何*or分配给列表的第一个链接&的songTitle元素；两者都是 type char*，所以这只是一个指针副本。跳过SongTitle和变量，它们没有任何作用AuthorName。SongNotes

Answer 2

SongTitle、AuthorName 和 SongNotes 这三个变量是 的数组char，而不是 的数组string。您需要将他们的声明更改为：

char* SongTitle[songCount];
char* AuthorName[songCount];
char* SongNotes[songCount];

然后，您需要像这样更新它们：

SongTitle[0] = songTitle0;//points at string songTitle0
AuthorName[0] = songArtist0;
SongNotes[0] = songNotes0;

当您将它们存储在链表中时：

    CurrentSong = malloc(sizeof(struct _song));
    CurrentSong->songTitle = SongTitle[tempCount];
    CurrentSong->songAuthor = AuthorName[tempCount];
    CurrentSong->songNote = SongNotes[tempCount];

Answer 3

这不是您应该使用链接列表的方式。

它是

typedef struct list {
    void *data;
    struct list *next;
}
SONG *s = (SONG *)songList->data;

同样，对于克隆字符串，您需要使用strdup.

例如

s->songTitle = strdup(SongTitle);
s->songAuthor = strdup(AuthorName);
s->songNote = strdup(SongNotes);

完成后不要忘记free琴弦。