Reverse integer bitwise without using loop

Question

I want to write a program which reverses the bits of an integer.
Ex 11000101 to 10100011
I know how to solve this using a loop, but I came across solutions that do it us

Answer 1

A recursive method for reversing the order of the bits in an integer -- call with the value to be reversed and the width of the value.

procedure int REVERSEBITS( int VALUE; int WIDTH ) {
  if WIDTH==1 then {
    return VALUE;
  } else {

    // intermediate values may help make the algorithm more understandable
    int HalfWidth = WIDTH >> 1;  // number of bits to be swapped at this level
    int HalfMask = HalfWidth-1;  // mask for left or right half of the value
    int RightHalfValue = VALUE & HalfMask;  // extract right half from value
    int LeftHalfValue = (VALUE >> HalfWidth) & HalfMask;  // extract left half

    // call reversing function on the two halves separately then swap the results
    return (REVERSEBITS(RightHalfValue, HalfWidth) << HalfWidth)
                | REVERSEBITS(LeftHalfValue, HalfWidth);
  }
}

Or, shorten the function by replacing all intermediate values with their definitions (optimizing compilers will usually produce the same code either way)

procedure int REVERSEBITS( int VALUE; int WIDTH ) {
  if WIDTH==1 then {
    return VALUE;
  } else {
    return (REVERSEBITS((VALUE & ((WIDTH>>1)-1)), (WIDTH>>1)) << (WIDTH>>1))
        | REVERSEBITS(((VALUE >> (WIDTH>>1)) & ((WIDTH>>1)-1)), (WIDTH>>1));
  }
}

Simplify the function code slightly by changing the functional definition, i.e. call with the value to be reversed and HALF the width of the value

procedure int REVERSEBITS( int VALUE; int WIDTH ) {
  if WIDTH==0 then {
    return VALUE;
  } else {
    return (REVERSEBITS((VALUE & (WIDTH-1)), (WIDTH >> 1)) << WIDTH)
        | REVERSEBITS(((VALUE >> WIDTH) & (WIDTH-1)), (WIDTH >> 1));
  }
}