Output of C programs | Set 64 (Pointers)
Prerequisite : Pointers in C
Question 1 : What will be the output of following program?
#include "stdio.h" int main() { char a[] = { 'A' , 'B' , 'C' , 'D' }; char * ppp = &a[0]; *ppp++; // Line 1 printf ( "%c %c " , *++ppp, --*ppp); // Line 2 } |
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?
#include <stdio.h> int main() { int * ptr; *ptr = 5; printf ( "%d" , *ptr); return 0; } |
OPTIONS:
a) compilation error
b) Runtime error
c) 5
d) linker error
OUTPUT: (b) Runtime error
Explanation: Pointer variable (*ptr) cannot be initialized.
Question 3 : What will be the output of following program?
#include <stdio.h> int main() { int a = 36; int * ptr; ptr = &a; printf ( "%u %u" , *&ptr, &*ptr); return 0; } |
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?
#include <stdio.h> int main() { int num = 10; printf ( "num = %d addresss of num = %u" , num, &num); num++; printf ( "\n num = %d addresss of num = %u" , num, &num); return 0; } |
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
OUTPUT: (d)
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?
#include <stdio.h> int main() { int i = 25; int * j; int ** k; j = &i; k = &j; printf ( "%u %u %u " , k, *k, **k); return 0; } |
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.
Recommended Posts:
- Output of C++ programs | Set 39 (Pointers)
- Output of C++ programs | Set 38 (Pointers)
- Output of C++ programs | Set 47 (Pointers)
- Output of C programs | Set 31 (Pointers)
- Output of C++ programs | Set 45
- Output of C programs | Set 55
- Output of C programs | Set 32
- Output of C++ Programs | Set 49
- Output of C programs | Set 52
- Output of C programs | Set 48
- Output of C programs | Set 54
- Output of C programs | Set 51
- Output of C programs | Set 38
- Output of C programs | Set 63
- Output of C programs | Set 63
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : BhanuSashankReddy