Bit exchange operation

I have been working on a certain task these days and after several hours of torture, I have done it! However, I believe the code that I have "crafted" is quite difficult to be understood and to be honest, I wouldn't be able to understand it if I wasn't the one who made it. Anyways, this is my code. I would like to receive some advice about what I could do to make it a little bit more simple and, by any chance, shorter:

```
using System;
class BitExchangeAdvanced
{
static void Main()
{
int num1, num2, p, q, k, loop;
uint n, n1, bit;
while (true)
{
// input number

Console.WriteLine("Input n:");
if (uint.TryParse(Console.ReadLine(), out n))
{
if (n >= 0)
{
break;
}
}
}

// input first position
while (true)
{
Console.WriteLine("Input p:");
if (int.TryParse(Console.ReadLine(), out p))
{
if (p >= 0)
{
break;
}
}
}

// input second position

while (true)
{
Console.WriteLine("Input q:");
if (int.TryParse(Console.ReadLine(), out q))
{
if (q >= 0)
{
break;
}
}
}

// input the limit
while (true)
{
Console.WriteLine("Input k:");
if (int.TryParse(Console.ReadLine(), out k))
{
if (k > 0)
{
break;
}
}
}

// check which position is first
if (p>q)
{
num1 = q;
num2 = p;
}
else
{
num1 = p;
num2 = q;
}

// the loop is used in the "for" loop
loop = num1;

// check if the inpputs are valid for completing the operation
if (num2 - num1 32 || num2 + k > 32)
{
Console.WriteLine("Overlapping or out of range");
}
else
{

// use "for" in order to complete the form

I think this is suggesting roughly (maybe exactly) the same technique as @harold does in his answer, but I'm not entirely certain that I understand his answer, so I can't say with certainty whether it's the same or not.

I'd consider two facts in coming to an answer. First, that where two bits are identical, we don't need to do anything to swap them--i.e., swapping a 1 with a 1 or a 0 with a 0 doesn't change anything.

Therefore, we only need to do anything where the bits are not equal to each other. In this case, we have to flip both bits--i.e., the one that was a 0 has to become a 1, and the one that was a 1 has to become a 0.

At least to me, that immediately brings an exclusive-or to mind. An exclusive-or produces a 1 if one input or the other (but not both) is a 1. Looking at it slightly differently, it produces a 1 if the bits are not equal to each other, and a 0 if they are equal to each other. Looked at from yet a slightly different angle, an XOR with 0 leaves a bit unchanged, and an XOR with 1 always flips the bit (changes a 0 to a 1, or a 1 to a 0).

With that in hand, the job becomes fairly simple: we start by XORing the bits in the two ranges. This gives us 0s where they contain identical bits (meaning we want to leave them unchanged). It gives us 1s where they contain differing bits (meaning we want to flip the bits in both ranges).

We then XOR that mask with the bits in each range. That leaves the bits in each range unchanged where they were identical to start with, and flips the bits in each range where they were different.

For example, let's consider swapping the first 4 bits of a byte with the second 4 bits:

first four: 0 1 0 1
last four:  0 1 1 0

When we XOR these, we get:

mask:       0 0 1 1

We then XOR that back with each of the input ranges:

first four: 0 1 0 1
mask:       0 0 1 1
result:     0 1 1 0

last four: 0 1 1 0
mask:      0 0 1 1 
result:    0 1 0 1

And now the first four have the same values as the last four did to start with (and vice versa).

Looking more specifically at your code, it has a fair number of places I think it could be made simpler and probably shorter. You start by reading four numbers, each with a block of code like this:

while (true)
{
 // input number

    Console.WriteLine("Input n:");
    if (uint.TryParse(Console.ReadLine(), out n))
    {
        if (n >= 0)
        {
            break;
        }
    }
}

The only things that change from one of these to the next are the string you display and the identity of the variable you're reading. This is an ideal candidate for being turned into a function. The function needs to read a number (just as you've done it above) and return the value when it's successful. Then the code in main will just call that function for each of the numbers it needs to read:

n = read_number("Input n:");
p = read_number("Input p:");
q = read_number("Input q:");
k = read_number("Input k:");

It's probably possible to avoid having to pass the string as well--you'd pass the variable itself, and figure out the name of that variable using reflection to build your string. That takes some rather more advanced use of .NET though, and I wouldn't advise it for now.

When it comes to finding which range is first:

if (p>q)
{
    num1 = q;
    num2 = p;
}
else
{
    num1 = p;
    num2 = q;
}

I think I'd just use p and q, but swap them if they're out of order:

if (p>q) {
    temp = p;
    p = q;
    q = temp;
}

(...and wish you were using C++, so you didn't have to re-implement basic operations like this for the 4 billionth time).

I've already advised a complete rewrite of the swapping code itself, so I guess there's not much point in reviewing that part of your code.

StackExchange Code Review Q#46469, answer score: 6