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I'm having issues converting 12 into hamming code. I've verified the that it's correctly converting the number into binary (base-2), and now when I attempt to use the formula to convert the security bits correctly, it's giving me a different answer than what I should be getting.


#include <iostream>
#include <vector>
using namespace std;

void print(vector<int> x){
   
    for (int i = 0; i < x.size(); i++){
        cout << x[i];
    }
    cout << endl;
  
}

vector<int> correctCode(vector<int> x, int pos){
    vector<int> correctedCode = x;
    cout << "Wrong Code: ";
    print(x);
    
    if (correctedCode[pos] == 0){
        correctedCode[pos] = 1;
    }
    else
    {
        correctedCode[pos] = 0;
    }
    cout << "Corrected Code: ";
    print(correctedCode);
    return correctedCode;
}

//Convert
void convertBinary(int val){
    vector<int> binaryVal;
    vector<int> binaryBase = {64,32,16,8,4,2,1};
    for (int i=0; i < 7; i++){
        if(val >= binaryBase[i]){
            val -= binaryBase[i];
            binaryVal.push_back(1);
        }
        else{
            binaryVal.push_back(0);
        }
    }
    //Security 1
    binaryVal[0] = (binaryVal[2]+binaryVal[4]+binaryVal[6])%2;

    
    //Security 2
    binaryVal[1] = (binaryVal[2]+binaryVal[5]+binaryVal[6])%2;
    
    //Security 3
    binaryVal[3] = (binaryVal[4]+binaryVal[5]+binaryVal[6])%2;
   
    
    
    print(binaryVal);
}

int main(int argc, const char * argv[]) {
    

    cout << "12 = ";
    convertBinary(12);
    cout << endl;

   
}

Binary = 0001100

12 = 1001100(Is what I'm getting) Should be = 0111100 (from what my professor mentioned)

Urban-Programmer
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    The programmer's secret weapon is the debugger. With a debugger you can run your program line by line and watch exactly what happens as it happens. Typical use is step through the program and keep an eye out for the unexpected. The unexpected is usually a bug. – user4581301 Jul 30 '21 at 20:13

1 Answers1

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I made an error when converting binary to hamming. I just assigned the last four digits of the binary code to message bits, then inserted the message bits after properly calculating the security bits.

    
    int m3,m5,m6,m7;
    m3 = binaryVal[3];
    m5 = binaryVal[4];
    m6 = binaryVal[5];
    m7 = binaryVal[6];
    
    binaryVal[0] = (m3+m5+m7)%2;
    binaryVal[1] = (m3+m6+m7)%2;
    binaryVal[3] = (m5+m6+m7)%2;
    
    binaryVal[2] = m3;
    binaryVal[4] = m5;
    binaryVal[5] = m6;
    binaryVal[6] = m7;
Urban-Programmer
  • 1