如果之前检查了 null，怎么可能有 null defernce

Question

我有这个功能可以搜索跳过列表中的元素。我不明白address sanitaizer给出的错误：

==5461==ERROR: AddressSanitizer: SEGV on unknown address 0x000000000010 (pc 0x555555555e28 bp 0x7fffffffdeb0 sp 0x7fffffffde90 T0)
==5461==The signal is caused by a READ memory access.
==5461==Hint: address points to the zero page.
    #0 0x555555555e28 in search_skip_list (/home/matteo/Scrivania/Algo/laboratorio-algoritmi-2021-2022-main/Esercizio 2/ex2/build/main+0x1e28)
    #1 0x5555555556fb in main (/home/matteo/Scrivania/Algo/laboratorio-algoritmi-2021-2022-main/Esercizio 2/ex2/build/main+0x16fb)
    #2 0x7ffff73c3d8f in __libc_start_call_main ../sysdeps/nptl/libc_start_call_main.h:58
    #3 0x7ffff73c3e3f in __libc_start_main_impl ../csu/libc-start.c:392
    #4 0x5555555552e4 in _start (/home/matteo/Scrivania/Algo/laboratorio-algoritmi-2021-2022-main/Esercizio 2/ex2/build/main+0x12e4)

AddressSanitizer can not provide additional info.
SUMMARY: AddressSanitizer: SEGV (/home/matteo/Scrivania/Algo/laboratorio-algoritmi-2021-2022-main/Esercizio 2/ex2/build/main+0x1e28) in search_skip_list
==5461==ABORTING
[Inferior 1 (process 5461) exited with code 01]

我的代码中没有发现分段错误。我是C新手，不知道如何正确使用gdb来查找问题。我把功能和结构初始化的方式放在这里，完整的代码太长，项目被文件占用。

void* search_skip_list(SkipList *list, void* item){
    if(list == NULL || item == NULL ) return NULL;

    Node *x = list->head;
    
    for (int i = list->max_level-1; i >= 0; i--)
    {   
        while (x->next[i]!=NULL && strcmp(item,x->next[i]->item) < 0)
        {
           x = x->next[i];
        }  
     }
    x = x->next[0];

    if(strcmp(item,x->item) == 0) return x->item;
    else{
        return "failure";
    } 
}

struct _SkipList {
    Node *head;
    unsigned int max_level;
    int (*compare)(void*, void*);
};
struct _Node {
    Node **next;
    unsigned int size;
    void *item;
};

SkipList* create_skip_list(){
    SkipList *list = malloc(sizeof(SkipList));
    list->max_level = 0;
    list->compare = NULL;
    list->head = create_head_node(NULL,MAX_HEIGHT);
    return list;
}

Node* create_head_node(void* item, int level){
    if(level <1) return NULL;
    Node *node = malloc(sizeof(Node));
    if(node == NULL){
        printf("error malloc node");
    }
    node->item = item;
    node->size = level;
    node->next = (Node**)malloc(level * sizeof(Node));
    for (int i = 0; i < node->size; i++)
        {
            node->next[i] = NULL;
        }
    if(node == NULL){
        printf("error malloc node->next");
    }
    return node;
}

我发现这可能是对 NULL 指针的尊重，但我不明白这怎么可能。但我认为这很奇怪，因为我首先检查是否有 NULL 值。还有其他问题可能会导致此错误吗？如何正确使用 GBD 找到问题所在的确切行？

我在函数之前使用断点运行 gdb，并且似乎在它第一次进入函数时停止，就好像第一个元素是 NULL 并且遵循 NULL 指针一样。

Answer 1

为了防止内存访问错误，我建议你这样重写你的函数：

Node* create_head_node(void* item, int level){
    if(level <1)
        return NULL;

    Node *node = malloc(sizeof(Node));
    if(node == NULL){
        printf("error malloc node
");
        /* Returning here prevent the program from accessing non allocated
         * memory. */
        return NULL;
    }

    node->item = item;
    node->size = level;

    node->next = (Node**)malloc(level * sizeof(Node *));
    if (!node->next) {
        printf("error malloc node next
");
        free(node);
        return NULL;
    }

    for (int i = 0; i < level; i++)
    {
        node->next[i] = NULL;
    }

    return node;
}