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- Question
What will be the output of the program?
#define P printf("%d\n", -1^~0); #define M(P) int main()\ {\ P\ return 0;\ } M(P)
Options- A. 1
- B. 0
- C. -1
- D. 2
- Correct Answer
- 0
- 1. What will be the output of the program?
#include<stdio.h> int main() { printf("%d >> %d %d >> %d\n", 4 >> 1, 8 >> 1); return 0; }
Options- A. 4 1 8 1
- B. 4 >> 1 8 >> 1
- C. 2 >> 4 Garbage value >> Garbage value
- D. 2 4 Discuss
- 2. If an unsigned int is 2 bytes wide then, What will be the output of the program?
#include<stdio.h> int main() { unsigned int m = 32; printf("%x\n", ~m); return 0; }
Options- A. ffff
- B. 0000
- C. ffdf
- D. ddfd Discuss
- 3. If an unsigned int is 2 bytes wide then, What will be the output of the program?
#include<stdio.h> int main() { unsigned int a=0xffff; ~a; printf("%x\n", a); return 0; }
Options- A. ffff
- B. 0000
- C. 00ff
- D. ddfd Discuss
- 4. What will be the output of the program?
#include<stdio.h> int main() { int i=32, j=0x20, k, l, m; k=i|j; l=i&j; m=k^l; printf("%d, %d, %d, %d, %d\n", i, j, k, l, m); return 0; }
Options- A. 0, 0, 0, 0, 0
- B. 0, 32, 32, 32, 32
- C. 32, 32, 32, 32, 0
- D. 32, 32, 32, 32, 32 Discuss
- 5. What will be the output of the program?
#include<stdio.h> int main() { int i=4, j=8; printf("%d, %d, %d\n", i|j&j|i, i|j&&j|i, i^j); return 0; }
Options- A. 4, 8, 0
- B. 1, 2, 1
- C. 12, 1, 12
- D. 0, 0, 0 Discuss
- 6. What will be the output of the program?
#include<stdio.h> int main() { printf("%d %d\n", 32<<1, 32<<0); printf("%d %d\n", 32<<-1, 32<<-0); printf("%d %d\n", 32>>1, 32>>0); printf("%d %d\n", 32>>-1, 32>>-0); return 0; }
Options- A. Garbage values
- B. 64 32
0 32
16 32
0 32 - C. All zeros
- D. 8 0
0 0
32 0
0 16
Discuss
- 7. Assuming a integer 2-bytes, What will be the output of the program?
#include<stdio.h> int main() { printf("%x\n", -1<<3); return 0; }
Options- A. ffff
- B. fff8
- C. 0
- D. -1 Discuss
- 8. What will be the output of the program?
#include<stdio.h> int main() { unsigned char i = 0x80; printf("%d\n", i<<1); return 0; }
Options- A. 0
- B. 256
- C. 100
- D. 80 Discuss
- 9. Bitwise & and | are unary operators
Options- A. True
- B. False Discuss
- 10. Left shifting a number by 1 is always equivalent to multiplying it by 2.
Options- A. True
- B. False Discuss
Bitwise Operators problems
Search Results
Correct Answer: 2 >> 4 Garbage value >> Garbage value
Correct Answer: ffdf
Correct Answer: ffff
Correct Answer: 32, 32, 32, 32, 0
Correct Answer: 12, 1, 12
Correct Answer: 64 32
0 32
16 32
0 32
Correct Answer: fff8
Explanation:
The system will treat negative numbers in 2's complement method.
Example:
Assume the size of int is 2-bytes(16 bits). The integer value 1 is represented as given below:
Binary of 1: 00000000 00000001 (this is for positive value of 1)
1's complement of binary 1: 11111111 11111110
2's complement of binary 1: 11111111 11111111
Thy system will store '11111111 11111111' in memory to represent '-1'.
If we do left shift (3 bits) on 11111111 11111111 it will become as given below:
11111111 11111111 ---(left shift 3 times)---> 11111111 11111000.
So, 11111111 11111000 ---(binary to hex)---> FF F8. (Required Answer)
Note:
How is the negative number obtained from 2's complement value?
As stated above, -1 is represented as '11111111 11111111' in memory.
So, the system will take 2's complement of '11111111 11111111' to the get the original negative value back.
Example:
Bit Representation of -1: 11111111 11111111
Since the left most bit is 1, it is a negative number. Then the value is
1's complement: 00000000 00000000
2's complement: 00000000 00000001 (Add 1 to the above result)
Therefore, '00000000 00000001' = 1 and the sign is negative.
Hence the value is -1.
Example:
Assume the size of int is 2-bytes(16 bits). The integer value 1 is represented as given below:
Binary of 1: 00000000 00000001 (this is for positive value of 1)
1's complement of binary 1: 11111111 11111110
2's complement of binary 1: 11111111 11111111
Thy system will store '11111111 11111111' in memory to represent '-1'.
If we do left shift (3 bits) on 11111111 11111111 it will become as given below:
11111111 11111111 ---(left shift 3 times)---> 11111111 11111000.
So, 11111111 11111000 ---(binary to hex)---> FF F8. (Required Answer)
Note:
How is the negative number obtained from 2's complement value?
As stated above, -1 is represented as '11111111 11111111' in memory.
So, the system will take 2's complement of '11111111 11111111' to the get the original negative value back.
Example:
Bit Representation of -1: 11111111 11111111
Since the left most bit is 1, it is a negative number. Then the value is
1's complement: 00000000 00000000
2's complement: 00000000 00000001 (Add 1 to the above result)
Therefore, '00000000 00000001' = 1 and the sign is negative.
Hence the value is -1.
Correct Answer: 256
Correct Answer: False
Explanation:
Bitwise & and | are not unary operators only bitwise ! is unary operator.
Correct Answer: True
Explanation:
0001 => 1
0010 => 2
0100 => 4
1000 => 8
0010 => 2
0100 => 4
1000 => 8
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