how to set in a binary form of number at some position 1 or 0

Suppose I have an eight bit number, I want to set at each bit position number 1 or zero, it is dynamic situation.

Suppose for example such situation, user enters two numbers which are equal or differs only by one, and I want that at each iteration from 0 position to seven, write these 0 and 1 in binary form of number, how can I implement it in cycle? Please help me.

An example:

int result = 0;

for (int i = 0; i < 8; i++) {
    int x, y;
    cin >> x >> y;

    if (x == y) {
        // set at i position 0;
    }

    else if ((x-y) == 1) {
        // set  at i position 1;(in result number)
    }
}

updated : it is what i want to implement : Adding two 8-bit two's complement binary numbers here is code for this

#include <iostream>
using namespace std;
int main(){
          int x,y;
          cin>>x>>y;
          int result=0;
          int carry=0;
         int sum=0;
          for (int i=0;i<8;i++){
              sum=carry;
           sum+= (x&(1<<i));
           sum+=(y&(1<<i));
              if (sum>1){
               sum-=2;
               carry=开发者_JS百科1;
              }
              else{



              carry=0;
              }
              result|=sum;
              result<<=1;



          }

           cout<<result<<" "<<endl;








 return 0;
}

You can change individual bits with AND and OR binary operators.

For example:

//set first bit to 1
value |= 1;

//set fourth bit to 0
value &= ~(1<<3);

//set 6th bit to 1
value |= (1<<5);

I don't know what happens if your inputs are different by two but you might want something like this:

int result = 0;

for (int i = 0; i < num_bits; ++i) {
    int a, b;
    std :: cin >> a >> b;

    result |= (a != b);
    result <<= 1;
}

Consider bit shifting.

To set the bit:

result |= (1<<i);

Un-setting the bit is left as an excercise to the reader.