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- Question
What is the output of the program?
#include<stdio.h> int main() { union a { int i; char ch[2]; }; union a u; u.ch[0] = 3; u.ch[1] = 2; printf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i); return 0; }
Options- A. 3, 2, 515
- B. 515, 2, 3
- C. 3, 2, 5
- D. None of these
- Correct Answer
- 3, 2, 515
Explanationprintf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i); It prints the value of u.ch[0] = 3, u.ch[1] = 2 and it prints the value of u.i means the value of entire union size.
So the output is 3, 2, 515.
More questions
- 1. What will be the output of the program?
#include<stdio.h> int main() { char *str; str = "%d\n"; str++; str++; printf(str-2, 300); return 0; }
Options- A. No output
- B. 30
- C. 3
- D. 300 Discuss
Correct Answer: 300
- 2. What will be the output of the program?
#include<stdio.h> #define SQUARE(x) x*x int main() { float s=10, u=30, t=2, a; a = 2*(s-u*t)/SQUARE(t); printf("Result = %f", a); return 0; }
Options- A. Result = -100.000000
- B. Result = -25.000000
- C. Result = 0.000000
- D. Result = 100.000000 Discuss
Correct Answer: Result = -100.000000
Explanation:
The macro function SQUARE(x) x*x calculate the square of the given number 'x'. (Eg: 102)Step 1: float s=10, u=30, t=2, a; Here the variable s, u, t, a are declared as an floating point type and the variable s, u, t are initialized to 10, 30, 2.
Step 2: a = 2*(s-u*t)/SQUARE(t); becomes,
=> a = 2 * (10 - 30 * 2) / t * t; Here SQUARE(t) is replaced by macro to t*t .
=> a = 2 * (10 - 30 * 2) / 2 * 2;
=> a = 2 * (10 - 60) / 2 * 2;
=> a = 2 * (-50) / 2 * 2 ;
=> a = 2 * (-25) * 2 ;
=> a = (-50) * 2 ;
=> a = -100;
Step 3: printf("Result=%f", a); It prints the value of variable 'a'.
Hence the output of the program is -100
- 3. What will be the output of the program?
#include<stdio.h> int main() { int i=4, j=-1, k=0, w, x, y, z; w = i || j || k; x = i && j && k; y = i || j &&k; z = i && j || k; printf("%d, %d, %d, %d\n", w, x, y, z); return 0; }
Options- A. 1, 1, 1, 1
- B. 1, 1, 0, 1
- C. 1, 0, 0, 1
- D. 1, 0, 1, 1 Discuss
Correct Answer: 1, 0, 1, 1
Explanation:
Step 1: int i=4, j=-1, k=0, w, x, y, z; here variable i, j, k, w, x, y, z are declared as an integer type and the variable i, j, k are initialized to 4, -1, 0 respectively.Step 2: w = i || j || k; becomes w = 4 || -1 || 0;. Hence it returns TRUE. So, w=1
Step 3: x = i && j && k; becomes x = 4 && -1 && 0; Hence it returns FALSE. So, x=0
Step 4: y = i || j &&k; becomes y = 4 || -1 && 0; Hence it returns TRUE. So, y=1
Step 5: z = i && j || k; becomes z = 4 && -1 || 0; Hence it returns TRUE. So, z=1.
Step 6: printf("%d, %d, %d, %d\n", w, x, y, z); Hence the output is "1, 0, 1, 1".
- 4. Which standard library function will you use to find the last occurance of a character in a string in C?
Options- A. strnchar()
- B. strchar()
- C. strrchar()
- D. strrchr() Discuss
Correct Answer: strrchr()
Explanation:
strrchr() returns a pointer to the last occurrence of character in a string.Example:
#include <stdio.h> #include <string.h> int main() { char str[30] = "12345678910111213"; printf("The last position of '2' is %d.\n", strrchr(str, '2') - str); return 0; }Output: The last position of '2' is 14.
- 5. A macro must always be defined in capital letters.
Options- A. True
- B. False Discuss
Correct Answer: False
Explanation:
FALSE, The macro is case insensitive.- 6. Point out the error in the program?
typedef struct data mystruct; struct data { int x; mystruct *b; };
Options- A. Error: in structure declaration
- B. Linker Error
- C. No Error
- D. None of above Discuss
Correct Answer: No Error
Explanation:
Here the type name mystruct is known at the point of declaring the structure, as it is already defined.- 7. Left shifting an unsigned int or char by 1 is always equivalent to multiplying it by 2.
Options- A. True
- B. False Discuss
Correct Answer: True
- 8. Can you combine the following two statements into one?
char *p; p = (char*) malloc(100);
Options- A. char p = *malloc(100);
- B. char *p = (char) malloc(100);
- C. char *p = (char*)malloc(100);
- D. char *p = (char *)(malloc*)(100); Discuss
Correct Answer: char *p = (char*)malloc(100);
- 9. How many bytes are occupied by near, far and huge pointers (DOS)?
Options- A. near=2 far=4 huge=4
- B. near=4 far=8 huge=8
- C. near=2 far=4 huge=8
- D. near=4 far=4 huge=8 Discuss
Correct Answer: near=2 far=4 huge=4
Explanation:
near=2, far=4 and huge=4 pointers exist only under DOS. Under windows and Linux every pointers is 4 bytes long.- 10. Can a structure can point to itself?
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
Explanation:
A structure pointing to itself is called self-referential structures.
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- 1. What will be the output of the program?