Assuming a integer 2-bytes, What will be the output of the program?
#include<stdio.h> int main() { printf("%x\n", -1<<3); return}
Which of the following statements are correct about the program?
#include<stdio.h> char *fun(unsigned int num, int base); int main() { char *s; s=fun(2); s=fun(16); printf("%s\n",s); return} char *fun(unsigned int num, int base) { static char buff[33]; char *ptr = &buff[sizeof(buff)-1]; *ptr = '\0'; do { *--ptr = "0123456789abcdef"[num %base]; num /=base; }while(num!=0); return ptr; }
On left shifting, the bits from the left are rotated and brought to the right and accommodated where there is empty space on the right?
Bitwise & and | are unary operators
What will be the output of the program?
#include<stdio.h> int main() { char c= int i, mask= for(i=i<=i++) { printf("%c", c|mask); mask = mask<< } return}
Which bitwise operator is suitable for checking whether a particular bit is on or off?
What will be the output of the program ?
#include<stdio.h> int main() { int i=j= printf("%d, %d, %d\n", i|j&j|i, i|j&&j|i, i^j); return }
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}
#include<stdio.h> int main() { unsigned int m[] = {0x0x0x0x0x0x0x0x80}; unsigned char n, i; scanf("%d", &n); for(i=i<=i++) { if(n & m[i]) printf("yes"); } return }
Bitwise & can be used to check if more than one bit in a number is on.
#define P printf("%d\n", -1^~0); #define M(P) int main()\ {\ P\ return 0;\ } M(P)
#include<stdio.h> int main() { unsigned char i = 0x printf("%d\n", i<<1); return}
Bitwise & can be used to check if a bit in number is set or not.
#include<stdio.h> int main() { int i=j=0xk, l, m; k=i|j; l=i&j; m=k^l; printf("%d, %d, %d, %d, %d\n", i, j, k, l, m); return }
#include<stdio.h> int main() { printf("%d %d\n", 32<<32<<0); printf("%d %d\n", 32<<-32<<-0); printf("%d %d\n", 32>>32>>0); printf("%d %d\n", 32>>-32>>-0); return}
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