Thread: Invert and reverse bits.

Hello,

I need to invert and reverse the bits in a one byte value. For example:

00111111 original
11000000 inverted
00000011 reversed

Here is what I have. It works, but I was wondering if there is a better or faster way.

Code:

#include <iostream>
#include <iomanip>

using namespace std;

int invert(int temp);

int main()
{
        int a = 0x3f;
        cout << hex << a << endl;
        cout << hex << invert(a) << endl;

        return 0;
}



int invert(int temp)
{
        int res         = 0;          //Result

        temp = 0xff - temp;     // ones complement?

        for(int x = 7; x > 0 ; x--)
        {
                res = res | (temp & 0x01);
                temp    = temp  >> 1;
                res       = res     << 1;
        }
        res = res | (temp & 0x01);

        return res;
}

There's a ~ operator for this:

Code:

int x = 0xA5A5A5A5;
int y = ~x;

printf("%x:%x\n", x, y);

I didn't read the part about reversing... there's no easy way for that, although it would probably be easiest to use the << and >> operators. The following might work, I didn't test it.

Code:

x = 0x0F0F0F0F;
y = 0;
for(i=0; i<32; ++i) // All 32 bits
{
   y |= (x & 0x1);
   y << 1;
   x >> 1;
}

Ignore that last post. I should probably register so I can edit. I didn't think that through enough.

Anyway, this program will reverse the bits in a 32-bit number:

Code:

#include <cstdio>

int main(int argc, char *argv[])
{
   int x = 0xF0F0F0F0;
   int y = 0;
   int i;

   for(i=0; i<32; ++i)
   {
      y = y << 1;
      y |= (x & 0x1);
      x = x >> 1;
   }
   printf("Y = %x\n", y);

   return 0;
}

Man, am I dumb today. You already had all that...

Although the shifting of the output needs to be before the OR, so that the last time through it doesn't shift after the OR (the last bit in the output should remain in the last position).