Confusion in data types in a 2D array

Question

Here is a code which print the address of the first element of an 2D array followed by an addition of 1. Though all 4 base addresses are same but their addition with 1 is ob

Answer 1

If you compile this as C++, you can use the typeid operator from the typeinfo STL library to get this information. For example, running this

#include 
#include 

void main()
{
     int array[4][3];
     printf("array %s\n",typeid(array).name());   //of type int(*)[3]
     printf("array+1 %s\n",typeid(array+1).name());
     printf("&array %s\n",typeid(&array).name());   //....???
     printf("&array+1 %s\n",typeid(&array+1).name());
     printf("array[0] %s\n",typeid(array[0]).name());    //of type int*
     printf("array[0]+1 %s\n",typeid(array[0]+1).name());
     printf("&array[0] %s\n",typeid(&array[0]).name()); //....???
     printf("&array[0]+1 %s\n",typeid(&array[0]+1).name());
}  

Gave me the following results

array int [4][3]
array+1 int (*)[3]
&array int (*)[4][3]
&array+1 int (*)[4][3]
array[0] int [3]
array[0]+1 int *
&array[0] int (*)[3]
&array[0]+1 int (*)[3]

This explains why adding 1 can increase the pointer by 4 bytes (a single int), 12 bytes (int[3]) or by 48 bytes (int[4][3]).