The ground is Rotating!

Here’s a Rotation function to recycle bits from the ‘bit bucket’,

Code:

---

unsigned char rol(unsigned char val) {
  int highbit;
  if(val & 0x80) // 0x80 is the high bit only
    highbit = 1;
  else
    highbit = 0;
  // Left shift (bottom bit becomes 0):
  val <<= 1;
  // Rotate the high bit onto the bottom:
  val |= highbit;
  return val;
}

---

Questions:
1. Why use 0x80(128) to check for the high bit?
2. Can you please explain the use of the Rotation?
3. Is there any other ways to do the Rotation?

Thanks.

We normally do not do other people's homework. Make a drawing with the bits and think about how the function works. Think about what the binary rerpesentation of 0x80 is.

Actually I understand the overall work of the function. It ANDs val and 0x80 to find if there’s a 1 in that position in val. But what I don’t understand is 0x80. Because, if we for example pass a value like 909, which has a 1 in the 8th position it works. The fallen bit after shifting is added to the end. But if we pass a value like 550, which 0 in 8th position it doesn’t Rotate the fallen bit?

You get an unsigned char to your function, so the range of values is (usually) 0-255 or 0x00-0xFF. Hence it doesn't make sense to check for 909 nor 550.

Yves