Output of C programs | Set 64 (Pointers)
Prerequisite : Pointers in C Question 1 : What will be the output of following program?
OPTIONS: a)C B b)B A c)B C d)C A
OUTPUT: (d) C A
Explanation: Line 1 : Now, ppp points to next memory location i.e., index 1 of the character array. Line 2 : Firstly, –*ppp= –(*ppp) is executed and hence the value ‘B’ (which is in the index 1 position of the char array) gets decremented by 1(i.e., it becomes ‘A’)and it is sent for printing. Then *++ppp= *(++ppp) is executed which initially increments the pointer to the next element of the array and prints the value in that index number 2 which is ‘C’. Although –*ppp is executed first compared to *++ppp, the display will be shown in the order as we mentioned in the printf() function in line 2. Hence we get output as C A. Question 2 : What will be the output of following program?
OPTIONS: a) compilation error b) Runtime error c) 5 d) linker error
OUTPUT: c 5
Explanation: Pointer variable (*ptr) cannot be initialized. Question 3 : What will be the output of following program?
OPTIONS: a) Address Value b) Value Address c) Address Address d) Compilation error
OUTPUT: (c)Address Address
Explanation: & and * cancelled each other and display the address stored in a pointer variable ptr i.e) the address of a. Question 4 : What will be the output of following program?
OPTIONS: a) Compilation error b) num = 10 address of num = 2293436 num = 11 address of num = 2293438 c) num = 10 address of num = 2293436 num = 11 address of num = 2293440 d) num = 10 address of num = 2293436 num = 11 address of num = 2293436
Explanation: Here a variable num holds the value 10 and get its address as 2293436, then increment is done as num++ which uses the same address space to store the incremented value. Question 5 : What will be the output of following program?
OPTIONS: a) address address value b) address value value c) address address address d) compilation error
OUTPUT : (a)address address value
Explanation : Here a pointer variable *j hold the address of i and then another pointer variable *k hold the address of j. now k = address of j *k = address of i **k = value of i. This article is contributed by Abhishek kurmi.
Please Login to comment...