Question

int i,sum=0;

  int array[8]={//You decide this};

  int *a,*b,*c;

  a = array;

  b = &array[3];

  c = &array[5];

  c[-1] = 2;

  array[1] = b[1]-1;

  *(c+1) = (b[-1]=5) + 2;

  sum += *(a+2) + *(b+2) + *(c+2);

  printf("%d %d %d %d\n",

sum, array[b-a], array[c-a], *a-*b);

array[8]={ ?? };

what is array[8]?

Answer 1

Answer: Array is [1, 1, 5, 4, 2, 6, 7, 8 ]

Note1**: (if initial array ={1,2,3,4,5,6,7,8})

Note2: Please refer the below code and comments for the detailed explanation

#include <stdio.h>

int main(void) {
        int i,sum=0;

  int array[8]={1,2,3,4,5,6,7,8};

  int *a,*b,*c;

  a = array; // a will now point to base address of array

  b = &array[3]; // b will now point to 4th element of array 

  c = &array[5]; // c will now point to 6th element of array 

  c[-1] = 2;  
    // c point to 6th element of array. 
    //Hence, 5th element of array will be updated to 2
    //array={1,2,3,4,2,6,7,8}
    
  array[1] = b[1]-1;
    // array[1]=2-1=1 (as b[1]=2)
    //array={1,1,3,4,2,6,7,8}

  *(c+1) = (b[-1]=5) + 2;
    // b points to 4th element of array. hence3rd element of array will be 5 (as b[-1]=5)
    // c point to 6th element of array
    // Here c+1 will point to 7th element of array
    // Value of 7th element will be updated to 7;
    // array={1,1,5,4,2,6,7,8}
  sum += *(a+2) + *(b+2) + *(c+2);
    //sum= 5 + 6+ 8 = 19
  printf("%d %d %d %d\n",sum, array[b-a], array[c-a], *a-*b); 
  array[8] = 99 // in array "array", 8th index is outside the defined boundary. Hence this assignment will be ignored.

        return 0;
}

Note3:Refer Screenshot for code indentation

The Output of the above code upon its execution will be ->

19 4 6 -3

Note4: Array[8] is an position which is outside the defined range of array(since array is defined for 8 elements of int type having position 0 to 7). Hence array[8]={some value} will not be reflected in array.