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;

哈希表的每个条目都是链表。然后，在main中我创建一个结构的实例，该结构具有一个结构变量，它是一个哈希表：

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;
}

打印表行的for循环中的print语句给出了这个输出：

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

然而，一旦我将哈希表传递给函数，只有前三行似乎存在。这是功能：

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for循环中的print语句中得到了这个输出：

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

最后两行似乎不存在。

我过去常常遇到分段错误，但我不再（我不知道原因）。

当我将上述功能更改为：

时

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;
}

它运作得很好。

Answer 1

您正在尝试创建五个链接列表，但您只有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，重新排列初始化代码，因为chux在评论中建议删除其中一个神奇的数字：

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

将哈希表传递给函数

1 个答案: