为什么这段代码没有在C中分配内存? [英] Why this code doesn't allocate memory in C?

问题描述

更新的问题在这里

HashTable中的内存分配问题

我正在用C制作HashTable.这就是我所做的.我认为我走的路是正确的，但是当我尝试

main.c

HashTablePtr hash;
hash = createHashTable(10);
insert(hash, "hello");
insert(hash, "world");

HashTable.c

    HashTablePtr createHashTable(unsigned int capacity){
    HashTablePtr hash;
    hash = (HashTablePtr) malloc(sizeof(HashTablePtr));
    hash->size = 0;
    hash->capacity = capacity;
    ListPtr mylist = (ListPtr)calloc(capacity, sizeof(ListPtr)); /* WHY IT DOESN'T ALLOCATE MEMORY FOR mylist HERE?? */
    mylist->head = NULL;
    mylist->size = 0;
    mylist->tail = NULL;    
    hash->list = mylist;  
    return hash;

ListPtr是一个LinkedList ptr

List.h

typedef struct list List;
typedef struct list * ListPtr;

struct list {
    int size;
    NodePtr head;
    NodePtr tail;
};
...
...

HashTable.h

    typedef struct hashtable * HashTablePtr;
    typedef struct hashtable HashTable;
    struct hashtable {
        unsigned int capacity;
        unsigned int size;
        ListPtr *list;
        unsigned int (*makeHash)(unsigned int, void *);
    };
...
...

运行调试器时，没有看到内存分配给myList.在上面的示例中，我试图将其制成10个列表的数组.

请帮助我解决这个问题.

如果有帮助，我不是C方面的专家.

解决方案

我个人并不喜欢使用typedef，特别是当您是初学者时.我认为这可能部分使您感到困惑.最好避免这样的事情:

typedef struct hashtable * HashTablePtr;

使用许多typedef将使您的代码难以阅读，因为您还需要不断查找它们所引用的内容.

主要问题是您要为哈希表/列表指针的大小分配内存，而不是为它们受尊重的结构分配大小.我认为下面的代码很好地说明了这一点.您还需要检查分配是否有效.如果是malloc，calloc，realloc.等等.如果失败，它们将返回NULL.如果发生这种情况，并且您不检查这种情况，则会出现segfault错误，并且程序将崩溃.

还要遵循c99标准，并将所有变量声明放在函数的开头.

c99 std

malloc联机帮助页

struct hashtable *
createHashTable(unsigned int capacity){
    struct hashtable *hash;
    struct list *mylist;

    /* You want to allocate size of the hash structure not the size of a pointer. */
    hash = malloc(sizeof(struct hashtable)); 
    // always make sure if the allocation worked.
    if(hash == NULL){
        fprintf(stderr, "Could not allocate hashtable\n");
        return NULL;
    }

    hash->size = 0;
    hash->capacity = capacity;

    /* Unless you need the memory to be zero'd I would just use malloc here
     * mylist = calloc(capacity, sizeof(struct list)); */
    mylist = malloc(capacity * sizeof(struct list));
    if(mylist == NULL){
        fprintf(stderr, "Could not allocate list\n");
        free(hash); /* free our memory and handle the error*/
        return NULL;
    }

    mylist->head = NULL;
    mylist->size = 0;
    mylist->tail = NULL;    
    hash->list = mylist;

    return hash;
}

还记得在释放哈希表之前释放您的列表:

free(myhash->list);
free(myhash);

Updated question is here

Memory allocation problem in HashTable

I'm working on making a HashTable in C. This is what I've done. I think I'm going on a right path but when I'm trying to

main.c

HashTablePtr hash;
hash = createHashTable(10);
insert(hash, "hello");
insert(hash, "world");

HashTable.c

    HashTablePtr createHashTable(unsigned int capacity){
    HashTablePtr hash;
    hash = (HashTablePtr) malloc(sizeof(HashTablePtr));
    hash->size = 0;
    hash->capacity = capacity;
    ListPtr mylist = (ListPtr)calloc(capacity, sizeof(ListPtr)); /* WHY IT DOESN'T ALLOCATE MEMORY FOR mylist HERE?? */
    mylist->head = NULL;
    mylist->size = 0;
    mylist->tail = NULL;    
    hash->list = mylist;  
    return hash;

ListPtr is a LinkedList ptr

List.h

typedef struct list List;
typedef struct list * ListPtr;

struct list {
    int size;
    NodePtr head;
    NodePtr tail;
};
...
...

HashTable.h

    typedef struct hashtable * HashTablePtr;
    typedef struct hashtable HashTable;
    struct hashtable {
        unsigned int capacity;
        unsigned int size;
        ListPtr *list;
        unsigned int (*makeHash)(unsigned int, void *);
    };
...
...

When I run my debugger, I see no memory being allocated to myList. In above example, my attempt is to make it an array of 10 lists.

Please help me to solve this.

I'm not that expert in C, if that helps.

解决方案

Personally I am not a huge fan of using typedefs, especially when you are a beginner. I think that may be partially what is confusing you. You are better avoiding things like:

typedef struct hashtable * HashTablePtr;

Using to many typedefs will make your code harder to read since you will need to constantly look up what they are referring too.

The main problem is that you were allocating the memory for the size of a hashtable/list pointer and not for the size of their respected structures. I think the code below shows this well. You will also want to check if you allocations worked. If malloc, calloc, realloc. etc. fail they return NULL. If this happens and you do not check for this case you will get a segfault error and your program will crash.

Also follow the c99 standard, and put all of your variable declarations at the start of the function.

c99 std

malloc manpage

struct hashtable *
createHashTable(unsigned int capacity){
    struct hashtable *hash;
    struct list *mylist;

    /* You want to allocate size of the hash structure not the size of a pointer. */
    hash = malloc(sizeof(struct hashtable)); 
    // always make sure if the allocation worked.
    if(hash == NULL){
        fprintf(stderr, "Could not allocate hashtable\n");
        return NULL;
    }

    hash->size = 0;
    hash->capacity = capacity;

    /* Unless you need the memory to be zero'd I would just use malloc here
     * mylist = calloc(capacity, sizeof(struct list)); */
    mylist = malloc(capacity * sizeof(struct list));
    if(mylist == NULL){
        fprintf(stderr, "Could not allocate list\n");
        free(hash); /* free our memory and handle the error*/
        return NULL;
    }

    mylist->head = NULL;
    mylist->size = 0;
    mylist->tail = NULL;    
    hash->list = mylist;

    return hash;
}

Also remember to free you list before you free your hashtable:

free(myhash->list);
free(myhash);

这篇关于为什么这段代码没有在C中分配内存?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！