Convert to 6bit

Question

I am trying to tidy up this function which converts a 10bit value to 6 bit. I also need to be able to define the bit length of the input for when I use a higher resolution ADC:

BYTE ioGetADC (void)                
 {
  BYTE r;

  ConvertADC();              // Start Conversion
  while(BusyADC());              // Wait for completion
   {
    r = ( (ReadADC())/16);          // Read result and convert to 0-63 (returns 10bit right hand justified)
   }

  return r;
 }

Answer 1

Building on Joachim's answer, how about:

uint32_t dropBits(uint32_t x, uint8_t bitsIn, uint8_t bitsOut)
{
  return x / (1 << (bitsOut - bitsIn));
}

so, for instance if we call dropBits(1023, 10, 6) to scale the maximum value of a 10-bit integer into 6 bits, it will return 1023 / (1 << 4) which is 1023 / 16 i.e. 63, the maximum for a 6-bit value.

Of course, we can be tempted to help the compiler out since the denominator is a power of two:

return x >> (bitsOut - bitsIn);

This removes the division operator, doing it with a shift directly instead.

Note that this can only drop bits, it can't scale values into more bits.

Answer 2

The divisor you need is the difference between the input and output bits, raised to the power of two.