c - 将哈希表传递给函数时数据丢失

Question

我创建了一个哈希表:

typedef struct _linked_list_{
    struct _linked_list_ *next;
    char *disk_name;
    struct disk *disk_object;
} linked_list;


typedef struct _hash_table_ {
    int size;
    linked_list **table;
} hash_table;

其中哈希表的每个条目都是一个链表。然后，在 main 中，我创建一个结构实例，该结构具有一个哈希表结构变量:

int main() {
    health_monitor *starbucks;
    starbucks = malloc(sizeof(health_monitor));
    starbucks->id = "92838382";

    // THIS IS THE HASH TABLE!
    starbucks->disks_in_system = malloc(sizeof(hash_table));
    starbucks->disks_in_system->table = malloc(sizeof(linked_list));
    starbucks->disks_in_system->size = 5;

    //initializing the first 5 rows to be NULL

    starbucks->disks_in_system->table[0] = NULL;
    starbucks->disks_in_system->table[1] = NULL;
    starbucks->disks_in_system->table[2] = NULL;
    starbucks->disks_in_system->table[3] = NULL;
    starbucks->disks_in_system->table[4] = NULL;

    //Making sure that the table rows were created correctly and contain NULL
    int counter;
    for(counter=0; counter <5; counter++){
        printf("The table row is: %s\n", starbucks->disks_in_system->table[counter]);
    }
    //passing the hash table into explore_current_directory function
    explore_current_directory(starbucks->disks_in_system, data_directory);       
    return 0;
}

for 循环中打印表行的 print 语句给出以下输出:

The table is: (null)
The table is: (null)
The table is: (null)
The table is: (null)
The table is: (null)

但是，当我将哈希表传递给函数后，似乎只有前三行存在。这是函数:

int explore_current_directory(hash_table* hm, char* directory){
    DIR *dp;
    struct dirent *ep;
    char* current_directory;
    dp = opendir(directory);

    int counter;
    for(counter=0; counter <5; counter++){
        printf("The table row is: %s\n", hm->table[counter]);
    }

    return 0;
}

我从上面的 for 循环内的 print 语句得到以下输出:

The table row is: (null)
The table row is: (null)
The table row is: (null)

最后两行似乎不存在。

在那之后我曾经遇到过段错误，但现在不再出现了(我不知道为什么。)

问题是，当我将上面的函数更改为:

int explore_current_directory(hash_table* hm, char* directory){
    int counter;
    for(counter=0; counter <5; counter++){
        printf("The table row is: %s\n", hm->table[counter]);
    }

    return 0;
}

效果很好。

Answer 1

您正在尝试创建五个链接列表，但您只为其中一个malloc足够的内存:

starbucks->disks_in_system->table = malloc(sizeof(linked_list*));

将该行更改为

starbucks->disks_in_system->table = malloc(5 * sizeof(linked_list*));

或者，更好的是，按照 chux 在注释中建议的方式重新排列初始化代码，以删除其中一个魔数(Magic Number):

starbucks->disks_in_system->size = 5;
starbucks->disks_in_system->table = malloc ( starbucks->disks_in_system->size
                                           * sizeof(linked_list*) );