What is the difference between the following two assignments?
int main()
{
int a=10;
int* p= &a;
int* q = (int*)p; <-------------------------
int* r = (int*)&p; <-------------------------
}
I am very much confused about the behavior of the two declarations.
When should i use one over the other?
int* q = (int*)p;
Is correct, albeit too verbose. int* q = p is sufficient. Both q and p are int pointers.
int* r = (int*)&p;
Is incorrect (logically, although it might compile), since &p is an int** but r is a int*. I can't think of a situation where you'd want this.
#include <stdio.h>
int main()
{
int a = 10; /* a has been initialized with value 10*/
int * p = &a; /* a address has been given to variable p which is a integer type pointer
* which means, p will be pointing to the value on address of a*/
int * q = p ; /*q is a pointer to an integer, q which is having the value contained by p, * q--> p --> &a; these will be *(pointer) to value of a which is 10;
int * r = (int*) &p;/* this is correct because r keeping address of p,
* which means p value will be pointer by r but if u want
* to reference a, its not so correct.
* int ** r = &p;
* r-->(&p)--->*(&p)-->**(&p)
*/
return 0;
}
int main()
{
int a=10;
int* p= &a;
int* q = p; /* q and p both point to a */
int* r = (int*)&p; /* this is not correct: */
int **r = &p; /* this is correct, r points to p, p points to a */
*r = 0; /* now r still points to p, but p points to NULL, a is still 10 */
}
Types matter.
The expression p has type int * (pointer to int), so the expression &p has type int ** (pointer to pointer to int). These are different, incompatible types; you cannot assign a value of type int ** to a variable of type int * without an explicit cast.
The proper thing to do would be to write
int *q = p;
int **r = &p;
You should never use an explicit cast in an assignment unless you know why you need to convert the value to a different type.
