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    Question

    Consider the following code snippet. What is the output

    of the program?  #include < stdio.h > int main ( ) {     int arr[5] = {1, 2, 3, 4, 5};     int *ptr = arr;     ptr + = 2;     printf("%d", *ptr);     return 0; }
    A 1 Correct Answer Incorrect Answer
    B 2 Correct Answer Incorrect Answer
    C 3 Correct Answer Incorrect Answer
    D 4 Correct Answer Incorrect Answer
    E 5 Correct Answer Incorrect Answer

    Solution

    In the code, ptr initially points to the first element of the array arr. The line ptr += 2; moves the pointer two positions ahead, meaning ptr now points to the third element of the array, which is 3. Therefore, *ptr dereferences the pointer to print the value 3. Why Other Options are Wrong: a) 1 is the first element, but the pointer has moved past it. b) 2 is the second element, but ptr skips over it. d) 4 is the fourth element, but ptr stops at the third. e) 5 is the fifth element, beyond where ptr is pointing.

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