Discussion
What will be the output of the program?
#include <stdio.h>
int main()
{
int n = 12;
printf("%d %d", n & 1, n >> 1);
return 0;
}Answer: C
Correct answer: 0 6
Two independent expressions are printed.
n & 1 isolates the least significant bit, which is the parity of the number: it is 1 for odd values and 0 for even ones. Since \( 12 \) is 1100, the low bit is 0, so the first value printed is 0. This is the usual fast alternative to n % 2, and it behaves better on negative numbers, where % in C yields \(-1\) rather than 1.
n >> 1 shifts every bit one place right, discarding the low bit: 1100 becomes 0110, which is 6. That is integer division by two, \( 12 / 2 = 6 \).
So the output is 0 6. Note that the two operations are complementary - the shift keeps everything the mask throws away - which is why they appear together when a number is consumed bit by bit.
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