How to right shift bits in c++?

Question

I have a hex number 0x8F (10001111 in binary). I want to right shift that value, so the new one would be 0xC7 (11000111). I tried with:

unsigned char x = 0x8F;
x=x>>1; 

but instead of 0xC7 I got 0x47? Any ideas on how to do this?

Answer 1

That's because what you want is a "rotate right", not "shift right". So you need to adjust for the lowest bit "falling off":

 x = ((x & 1) << CHAR_BITS-1) | (x >> 1);

should do the trick.

[And at least some compilers will detect this particular set of operations and convert to the corresponding ror or rol instruction]

Answer 2

Right shifting or left shifting will fill with 0s respectively on the left or on the right of the byte. After shifting, you need to OR with the proper value in order to get what you expect.

x = (x >> 1); /* this is now 01000111 */
x = x | ( 0x80 ); /* now we get what we want */

Here I'm ORing with the byte 10000000 which is 0x80 resulting in 0xC7.

Making it more concise, it becomes:

x = (x >> 1) | (unsigned char)0x80;

Answer 3

Right shift on an unsigned quantity will make new zeros to enter, not ones.

Note that right shift is not a right rotation. To do that you need

x = (x >> 1) | (x << 7)