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Bitwise Operators
What will be the output of the program? #include
int main() { int i=4, j=8; printf("%d, %d, %d\n", i|j&j|i, i|j&&j|i, i^j); return 0; }
4, 8, 0
1, 2, 1
12, 1, 12
0, 0, 0
Correct Answer:
12, 1, 12
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Bitwise Operators
What will be the output of the program? #include
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; }
If an unsigned int is 2 bytes wide then, What will be the output of the program? #include
int main() { unsigned int a=0xffff; ~a; printf("%x\n", a); return 0; }
If an unsigned int is 2 bytes wide then, What will be the output of the program? #include
int main() { unsigned int m = 32; printf("%x\n", ~m); return 0; }
What will be the output of the program? #include
int main() { printf("%d >> %d %d >> %d\n", 4 >> 1, 8 >> 1); return 0; }
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)
What will be the output of the program? #include
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; }
Assuming a integer 2-bytes, What will be the output of the program? #include
int main() { printf("%x\n", -1<<3); return 0; }
What will be the output of the program? #include
int main() { unsigned char i = 0x80; printf("%d\n", i<<1); return 0; }
Bitwise & and | are unary operators
Left shifting a number by 1 is always equivalent to multiplying it by 2.
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