Why is it safer to use sizeof(*pointer) in malloc

Question

Given

struct node
{
     int a;
     struct node * next;
};

To malloc a new structure,

struct node *p = malloc(sizeof(*p));         


        

Answer 1

Because if at some later point in time p is made to point to another structure type then your memory allocation statement using malloc doesn't have to change, it still allocates enough memory required for the new type. It ensures:

• You don't have to modify the memory allocation statement every time you change the type it allocates memory for.
• Your code is more robust and less prone to manual errors.

In general it is always a good practice to not rely on concrete types and the the first form just does that ,it doesn't hard code a type.