Question no 7 in MCQ-Pointers

#1

Please explain question no 7

Q­7 In the piece of code, arr[ ][ ] is a 2­D array and assume that the
contents of the 2­D array are already filled up. What is stored in the
variable 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);

#2

Hey @slow_motion, In this code *(arr+i)+(i++) this will results in the addresses where i can be 0,1,2 in the loop and * ( * (arr+i)+(i++)) results in the garbage values at these addresses and then we are simply adding these values using the variable sum.

#3

Hope this would help :slight_smile:

#4

@mr.encoder No it doesn’t help. Can you please elaborate with an example.

#5

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.

1 Like
#6

Got it…thank you…

#7

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