Pointers MCQ problem How is sum storing sum of elements of principal diagonal?

#1

In the piece of code, arr[ ][ ] is a 2­D array and assume that thecontents of the 2­D array are already filled up. What is stored in th evariable sum at the end of the code segment?
int arr[3][3];
int i, sum=0; i = 0;
while(i<3) {sum += * ( * (arr+i)+(i++)); }
printf(“sum:%d”, sum);A) Sum of all elements in the matrix
B) Sum of alternate elements in the matrix
C) Sum of the elements along the principal diagonal
D) None

#2

@dsingh200021 hey darshpreet
for 7th question
int arr[3][3];
int i, sum=0;
i = 0;
while(i<3)
{
sum += ((arr+i)+(i++));
}
printf(“sum:%d”, sum);
Options:
Sum of all elements in the matrix
Sum of alternate elements in the matrix
Sum of the elements along the principal diagonal
None of these
Answer: Sum of the elements along the principal diagonal
Explanation:
Observe that the unary operator ++ in the program is a post increment operator. Thus
the value of i gets incremented only after the execution of the statement.
Since arr is a 2-D array, arr represents a pointer to an array of pointers, where each
pointer in the array in turn points to a 1-D array(i^th pointer points to the starting location
of i^th row in the array). So, *(arr+i) represents the starting address of i^th row. This
implies *(arr+i)+i represents the address of i^th element in the i^th row, which is
basically a diagonal element.

#3

Thanks for your reply but I am still confused
*(arr+i) represent the VALUE of ith element.
Lets have a 2d array {1,2
3,4}
Then *(arr+i) for i=0 should give value 1 and *(arr+i)+(i++) should give value 2

I WANT TO ASK IF *(ARR+I) GIVES THE VALUE OR ADDRESS because * is dereferencing pointer.

#4

You have this as your last second line I think it should give value at that address because a dereferencing operator * is used