c programming

Question

unsigned int num;
 int i;
 scanf("%u",&num);
 for(i=0;i<16;i++)
 {
     printf("%d",(num<<i & 1<<15)?1:0); // how it will execute?
 }

Answer 1

suppose u have entered num as 11 (in binary it will be 0000000000001011)

<< is left shift operator and <<  i left shifts by i every time       & is bitwise and operator returns 1 if both bits are 1 else return 0

1<<15 means left shift 1 in binary to left side by 15 places so it will be

0000000000000001<<15 =100000000000000

now for i=0

0000000000001011<<0 & 1000000000000000 =0 (with or without  ? 1:0

for i=1

000......10110 &1000000000000000 =0 (with or without  ? 1:0

and so on till i becomes 12 then it will retuen 1 as their is 1 on msb on both side of & operator

then 0 then 1 then 1 so ans will be

0000000000001011

so actually enetered number will be shown in its equvt binary equivalent

Comments

Thank you sir