Binary Set Calculator Function - c++

Question

So I've been working on a project for my C++ class, and we have to create a binary calculator. Yet the Professor says the functions should return an 8 bit binary set back. My issue is this
11111111 + 11111111 = 0111111110
Yet in the function we originally create this is the outcome
11111111 + 1111111 = 00000000
Which to me is incorrect. So I changed my function to this
Decimal to Binary
string DecToBin(int num)
{
/*
Purpose: Changing a Decimal to a Binary Set
Pre: Valid positive integer
Post: Returns the valid binary string
*/
string bin = "";
while (num >= 0)
{
bin += (num % 2 == 0 ? "0" : "1");
if (num != 0)
num /= 2;
else break;
}
return bin;
}
Though here lies the issue again
01010101 + 10101010 = 011111111
But my function above returns
01010101 + 10101010 = 111111110
What would be the best possible function to create if one I NEED to return an 8 bit set or if like my said function above which returns the correct answer for some and wrong for the others which I need to figure out why that is in the first place.
Binary to Decimal
int BinToDec(string bin)
{
/*
Purpose: To generate a decimal integer from a string binary set
Pre: Valid String binary set
Post: Output the output decimal integer
*/
int output = 0; //initialize output as 0
int base2Start = 128;//base2 start at 128
int len = bin.length();//get the string length
for (int i = 0; i < len; i++)//iterate
{
if (bin[i] == '1')//if bin[i] in the array of string is a char 1
{
output = output + base2Start;//output gets + base2Start
}//end if condition
base2Start = base2Start / 2;//divide base2Start after each iteration
}//end for loop
return output;//return the output
}
Addition Function
int Addition(string st1, string st2)
{
/*
Purpose: Get's two valid Binary sets, then adds their decimal conversion, and returns the addition
Pre: Need two strings that SHOULD be valid binary
Post: Returns binary from decimal conversion
*/
int first, second;
if (ValidBin(st1))
{
first = BinToDec(st1);
}
else return 0;
if (ValidBin(st2)){
second = BinToDec(st2);
}
else return 0;
add++;
return first + second;
}

Answer 1

bin += (num % 2 == 0 ? "0" : "1");
Should be
bin = (num % 2 == 0 ? "0" : "1") + bin;
Since you each time you are adding the least significant bit of the num to string. So at the end, according to your code, you will have the least significant most left instead of most right.
Edit: In order to truncate the result to 8 bit width change the following line:
return first + second;
By the this one:
return (first + second) & 0xFF; // Same as (first + second) % 256

Answer 2

#include <iostream>
using namespace std;
string sumBinary (string s1, string s2);
int main()
{
cout << "output = "<< sumBinary ("10","10");
}
string sumBinary (string s1, string s2)
{
if (s1.empty())
return s2;
if (s2.empty())
return s1;
int len1 = s1.length() -1;
int len2 = s2.length() -1;
string s3;
s3 = len1 > len2 ? s1: s2;
int len3 = s3.length() -1;
bool carry = false;
while (len1>=0 || len2>=0) {
int i1 = len1>=0? s1[len1--] - '0': 0;
int i2 = len2>=0? s2[len2--] - '0': 0;
// Check if any invalid character
if (i1 <0 || i1>1 || i2<0 || i2>1)
return "";
// 3 bit sum
int sum = i1 ^ i2 ^ carry;
// 3 bit carry
carry = (i1 & carry) | (i2 & carry) | (i1 & i2);
s3[len3--] = '0' + sum;
}
if (carry)
s3 = "1" + s3;
return s3;
}