C-无法释放在链表结构中分配的内存

Question

Consider the following code snippet

  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;
    }

Question: I am not able to free current->name, only when i comment the freeing of name, program works. If I uncomment the free part of current->name, I get Heap corruption error in my visual studio window. How can I free name ?

Reply:

@all,YES, there were typos in struct declaration. Should have been char* name, and struct node* next. Looks like the stackoverflow editor took away those two stars.

The issue was resolved by doing a malloc(nameLength + 1). However,If I try running the old code (malloc(namelength)) on command prompt and not on visual studio, it runs fine. Looks like, there are certain compilers doing strict checking.

One thing that I still do not understand is , that free does not need a NULL termination pointer, and chances to overwrite the allocated pointer is very minimal here.

user2531639 aka Neeraj

Answer 1

This is writing beyond the end of the allocated memory as there is no space for the null terminating character, causing undefined behaviour:

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

To correct:

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

Note calling free() with a NULL pointer is safe so checking for NULL prior to invoking it is superfluous:

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

---

Additionally, I assume there are typos in the posted struct definition (as types of name and next should be pointers):

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