Is bitwise complement of (n-1) equals to (-n)

Question

The following code prints the two odd occuring numbers in an array:

#include <iostream>
using namespace std;
int main(){
   int n;
   cin>>n;
   int arr[n],x=0;
   for(int i=0;i<n;i++){
      cin>>arr[i];
      x=x^arr[i];
   }
   x=(x&~(x-1));
   int reso=0,rest=0;
   for(int i=0;i<n;i++){
      if(x&arr[i])
         reso=reso^arr[i];
      else
         rest=rest^arr[i];
   }
   cout<<reso<<' '<<rest;
   return 0;
}

I observed that ~(x-1)==(-x) in 2's complement hence I replaced the ~(x-1) with (-x) and wrote the following code, which worked fine :

#include <iostream>
using namespace std;
int main(){
   int n;
   cin>>n;
   int arr[n],x=0;
   for(int i=0;i<n;i++){
      cin>>arr[i];
      x=x^arr[i];
   }
   x=(x&(-x)); //changed expression
   int reso=0,rest=0;
   for(int i=0;i<n;i++){
      if(x&arr[i])
         reso=reso^arr[i];
      else
         rest=rest^arr[i];
   }
   cout<<reso<<' '<<rest;
   return 0;
}

So I need help, is ~(x-1) is equal to (-x) in 2's complement or I am misconcluding here, kindly provide explanation ?

Answer 1

Taking the 1's complement of a d-1 digits number is the same as subtracting it from the number 111...1 (d-1) digits, which is 2^d-1.

E.g. with five bits numbers,

9d = 01001b; ~9d = 10110b = -16d + 6d = -10d

and

-9d = -16d + 6d = 11010b