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

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)

Answer 2

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 3

Right shifting or left shifting will fill with 0 s 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 OR ing with the byte 10000000 which is 0x80 resulting in 0xC7 .

Making it more concise, it becomes:

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