C - 无法释放链接列表结构中分配的内存

Question

请考虑以下代码段

  struct node {
    char *name;
    int m1;
    struct node *next;
    };

    struct node* head = 0; //start with NULL list

    void addRecord(const char *pName, int ms1)
    {   
        struct node* newNode = (struct node*) malloc(sizeof(struct node)); // allocate node

        int nameLength = tStrlen(pName);
        newNode->name = (char *) malloc(nameLength);
        tStrcpy(newNode->name, pName);

        newNode->m1 = ms1;
        newNode->next = head; // link the old list off the new node
        head = newNode;
    }

    void clear(void)
    {
        struct node* current = head;
        struct node* next;
        while (current != 0) 
        {
            next = current->next; // note the next pointer
    /*      if(current->name !=0)
            {
                free(current->name);
            }
    */
            if(current !=0 )
            {
                free(current); // delete the node
            }
            current = next; // advance to the next node
        }
        head = 0;
    }

问题:我无法释放当前 - >名称,只有当我评论释放名称时,程序才有效.如果我取消注释current-> name的免费部分,我的visual studio窗口中会出现Heap损坏错误.我怎么能免费提名？

答复:

@ all,YES,结构声明中有拼写错误.应该是char*name和struct node*next.看起来stackoverflow编辑器夺走了这两颗星.

通过执行malloc(nameLength + 1)解决了该问题.但是,如果我尝试在命令提示符下运行旧代码(malloc(namelength))而不是在visual studio上运行,它运行正常.看起来,某些编译器正在进行严格的检查.

我仍然不明白的一件事是,free不需要NULL终止指针,并且在这里覆盖分配指针的可能性非常小.

user2531639又名Neeraj

Answer 1

这是写入超出已分配内存的末尾,因为空终止字符没有空间,导致未定义的行为:

newNode->name = (char *) malloc(nameLength);
tStrcpy(newNode->name, pName);

纠正:

newNode->name = malloc(nameLength + 1);
if (newNode->name)
{
    tStrcpy(newNode->name, pName);
}

注意free()使用NULL指针调用是安全的,因此NULL在调用它之前检查是多余的:

free(current->name);
free(current);

另外,我假设在发布的struct定义中存在拼写错误(作为类型name和next应该是指针):

struct node {
    char* name;
    int m1;
    struct node* next;
};