Question

Here,I have some Doubt with the output.

Why the Output is same ?

   int (*r)[10];
   printf("r=%p  *r=%p\n",r,*r);
   return 0;

Platform- GCC UBUNTU 10.04

Answer 1

Because Name of the array decays to an pointer to its first element.

   int (*r)[10]; 

Is an pointer to an array of 10 integers.
r gives you the pointer itself.

This pointer to the array must be dereferenced to access the value of each element.
So contrary to what you think **r and not *r gives you access to the first element in the array.
*r gives you address of the first element in the array of integers, which is same as r

Important to note here that:
Arrays are not pointers

But expressions involving array name sometimes behave as pointer when those being used as name of the array would not make sense.