Output of following program?
# include <stdio.h> void fun(int *ptr) { *ptr = 30; } int main() { int y = 20; fun(&y); printf("%d", y); return 0; } |
(A) 20
(B) 30
(C) Compiler Error
(D) Runtime Error
Answer: (B)
Explanation: The function fun() expects a pointer ptr to an integer (or an address of an integer). It modifies the value at the address ptr. The dereference operator * is used to access the value at an address. In the statement ‘*ptr = 30’, value at address ptr is changed to 30. The address operator & is used to get the address of a variable of any data type. In the function call statement ‘fun(&y)’, address of y is passed so that y can be modified using its address.
Recommended Posts:
- C | C Quiz - 113 | Question 1
- C Quiz - 102 | Question 1
- C Quiz - 101 | Question 2
- C Quiz - 101 | Question 1
- C | Input and Output | Question 13
- C | Pointer Basics | Question 11
- C | Pointer Basics | Question 7
- C | Pointer Basics | Question 17
- C | Pointer Basics | Question 4
- C | Pointer Basics | Question 3
- C | Pointer Basics | Question 1
- C | Macro & Preprocessor | Question 2
- Output of C programs | Set 64 (Pointers)
- C | Macro & Preprocessor | Question 15
- Compiling a C program:- Behind the Scenes