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
This is writing beyond the end of the allocated memory as there is no space for the null terminating character, causing undefined behaviour:
To correct:
Note calling
free()
with aNULL
pointer is safe so checking forNULL
prior to invoking it is superfluous:Additionally, I assume there are typos in the posted
struct
definition (as types ofname
andnext
should be pointers):