How to
if (x == 1) printf("2\n");
else if (x == 2) printf("1\n");
else printf("0\n");
using bitwise operators?
My attempt is:
for (x = 0; x < 8; x++) {
printf("%d\n", 0x03 & (0x03 ^ x));
}
Output:
3
2
1
0
3
2
1
0
Desired output:
0
2
1
0
0
0
0
0
This is insane, but I finally figured it out:
printf("%d\n", (3 & ~(x & 3)) & (0xfc >> (x << 1)));
- is not a bitwise operator. Bonus points if you can replace it with a bitwise equivalent.
Not sure about the "bitwise" requirement ... If you want to compress the code, for some weird reason (the if version is very easy to understand which is a good thing), you might do something like this:
printf("%c\n", "210"[i == 0 ? 0 : i == 1 ? 1 : 2]);
This is of course almost the worst possible solution, since it's overly "clever": any reader of the code must spend valuable time to decode it, to understand what's going on.
Here is an answer that needs only two operations:
printf("%d\n", (4 >> x) & 3);
x < 0 or x >= sizeof(int) * CHAR_BIT.
Not sure why but...
for (x = 0; x < 8; x++) {
printf("%d %d\n", x, !x|~3&x?0:3^x);
}
After several false starts, a solution with six bitwise operations:
#include <stdio.h>
int main()
{
int x, y;
for (x = 0; x < 8; ++x)
{
y = !(x ^ 1) << 1 | !(x ^ 2);
printf("x = %d, y = %d\n", x, y);
}
return 0;
}
(Note that it's arguable as to whether ! is a true bitwise operator.)
printf("%d\n", x & 0xfc || !x ? 0 : 3 - x & 3);should do the job.x = 0