Progrmming

Question

int main()
{
    int a[5]={1,2,3,4,5};
    int *ptr=(int *)(&a+1); 
    // why is typecasting necessary here? It can just be int*ptr=&a+1;
    printf("%d%d",*(a+1),*(ptr-1));
}

• A. 2 5
• B. garbage
• C. compiler defined
• D. segmentation fault

Comments

Yes here no need of typecasting but iam not sure about answer I thought answer will be 21

---

&a indicates the entire array

&a+1 goes to the next location outside the array

now when it is decremented ,it comes to position where it's 5

and 5 gets printed

Answer 1

It is not necessary to type cast,actually there is no need to type cast here because &a+1 is representing address  which is of type integer only and int *ptr is storing the address of type integer only.

Answer 2

We always have to respect the "types" though C does not always enforce it. But internally it uses the type a lot-- especially when it comes to pointers. Those who thinks in terms of types can find pointers really easy. 

int *ptr=(int *)(&a+1); 

Here, a is an int array. a+1 points to the second element of the array. As pointer addition says, this is numerically a + sizeof(*a) - whatever be the type of 'a'. Now, when we do &a, it returns the address of a, and the type changes to pointer to an int array of size 5. So, now, when we do b + 1, (assume b = &a), it becomes b + sizeof(*b) = &a + sizeof(5 *int). So, this just goes to the next element outside the array. 

Now after addition also we get value whose type is pointer to an int array of 5 elements. And we are assigning it to a pointer to int. So, this requires a typecast. Otherwise do 

int (*p)[5] = &a+1;

Now, in the printf, *(a+1) is used. Prints the second element of array - 2. *(ptr-1) - prints the last element which is 5. 

Comments

Then for every pointer declaration there must be a typecast

eg. considering a to be an array of 5 elements

Int *p=&a;

this returns a pointer to an array of 5 elements  to a pointer to a int

so it must be

int *p= (int*)&a;

 

have I properly understood what you tried to convey

correct me if am wrong

 

and the above code executes well without the typecast too, so is it that typecasting is not mandatory but that's how it's handled?