c - 将哈希表传递给函数时数据丢失

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一个链表。然后，主要是创建一个结构实例，该实例的结构变量是哈希表：

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;
}

for循环内打印表行的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，一旦我将哈希表传递到函数中，似乎仅存在前三行。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语句获得以下输出：

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，我曾经遇到过细分错误，但现在不了（我不知道为什么）。

事实是，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;
}

它工作正常。

最佳答案

您正在尝试创建五个链表，但是您只能malloc足够一个表的内存：

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，按照注释中建议的chux重新排列初始化代码，以删除其中一个魔术数字：

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

关于c - 将哈希表传递给函数时数据丢失，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/31300430/