Ex 14: What is the output?

main( ){
 static int  a[ ]   = {0,1,2,3,4};
 int  *p[ ] = {a,a+1,a+2,a+3,a+4};
 int  **ptr =  p;
 ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *++ptr;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 ++*ptr;
       printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
}
Ans:
      111
      222
      333
      344
Explanation:

Let us consider the array and the two pointers with some address
a    
0     1     2     3     4
100      102      104      106      108
 p
100   102   104   106   108
  1000    1002    1004    1006    1008
  ptr   
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1,  *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102  so the value is (102 – 100)/(scaling factor) = 1,  **ptr is the value stored in the location pointed by  the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is  1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.