Question

Design a transducer to convert a binary string into octal. For example the bit string 001101110 should produce 156.

Please complete the code to account for the 7 cases of 3 digit binary strings.

//Function binaryToOctal takes a char array digits of length 3

//Pre: digits contains 3 binary digits.

//Post: function returns the octal number equivalent to the 3 binary digits

int binaryToOctal( char digits[], int 3){

int number;

if(digits[0]=='0')

if (digits[1]=='1')

if (digits[2]=='0')

return 2;//found "010"

else return 3;//found "011"

else if(digits[1]=='0')

if (digits[2]== '1'

return 1// found "001"

.......//keep listing all combinaisons until all are considered.

C++

Complete code

Use arr declared in main function for testing

#include <iostream>
using namespace std;
int binaryToOctal( char digits[], int a){
int number;
if(digits[0]=='0'){

        if (digits[1]=='1')
                if (digits[2]=='0')
                        return 2;//found "010"
                 else return 3;//found "011"
        else if(digits[1]=='0')
                if (digits[2]== '1')
                        return 1;// found "001"
                else return 0;//found "000"     
}
else{ 
if(digits[0]=='1')
    if (digits[1]=='1')
                if (digits[2]=='0')
                        return 6;//found "110"
                else return 7;//found "111"
        else if(digits[1]='0')
            if(digits[1]=='0')
                if (digits[2]== '1')
                        return 5;// found "101"
                else return 4;//found "100"
                }
                
}
int main(){
    char arr[3]={'1','0','0'};
int ans=binaryToOctal(arr,3);
cout<<ans;
                        
return 0;                       
}

Answer 1

Step1:

We will traverse the given array from back and take 3 values from array every time .

We will appropriately check the size left of array and append 0 as required as:

  //for three values
        if (i >= 2)
        {
            char temp[3] = {arr[i - 2], arr[i - 1], arr[i]};
            //call function for finding the octal in three
            int ans = binaryToOctal(temp, 3);
            //appending to new string  after converting the integer to char
            res += (ans + '0');
            i -= 3;
        }
        //but if only size left with 2 then add one 0 and take two element
        else if (i == 1)
        {
            char temp[3] = {0, arr[i - 1], arr[i]};
            //call function for finding the octal in three
            int ans = binaryToOctal(temp, 3);
            //appending to new string  after converting the integer to char
            res += (ans + '0');
            i -= 2;
        }
        //similarly add two 0 then one value for size 1 left
        else if (i == 0)
        {
            char temp[3] = {0, 0, arr[i]};
            //call function for finding the octal in three
            int ans = binaryToOctal(temp, 3);
            //appending to new string  after converting the integer to char
            res += (ans + '0');
            i--;
        }
        else
        {
            i--;
        }

Step2:

Appending to string after converting to integer as:

res += (ans + '0');

Step3:

Code: Evrything about code is explained in comments of code:

#include <iostream>
#include <algorithm>
using namespace std;
int binaryToOctal(char digits[], int a)
{
    int number;
    if (digits[0] == '0')
    {

        if (digits[1] == '1')
            if (digits[2] == '0')
                return 2; //found "010"
            else
                return 3; //found "011"
        else if (digits[1] == '0')
            if (digits[2] == '1')
                return 1; // found "001"
            else
                return 0; //found "000"
    }
    else
    {
        if (digits[0] == '1')
            if (digits[1] == '1')
                if (digits[2] == '0')
                    return 6; //found "110"
                else
                    return 7; //found "111"
            else if (digits[1] = '0')
                if (digits[1] == '0')
                    if (digits[2] == '1')
                        return 5; // found "101"
                    else
                        return 4; //found "100"
    }
}
int main()
{
    char arr[] = {'0', '0', '1', '1', '0', '1', '1', '1', '0'};
    int n = (sizeof(arr) / sizeof(arr[0]));
    cout << "For the input array : ";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    cout << endl;
    cout << "Octal is: ";
    int i = n - 1;
    string res = "";
    //traverse the loop from back and take three values from left to right
    while (i >= 0)
    {
        //for three values
        if (i >= 2)
        {
            char temp[3] = {arr[i - 2], arr[i - 1], arr[i]};
            //call function for finding the octal in three
            int ans = binaryToOctal(temp, 3);
            //appending to new string  after converting the integer to char
            res += (ans + '0');
            i -= 3;
        }
        //but if only size left with 2 then add one 0 and take two element
        else if (i == 1)
        {
            char temp[3] = {0, arr[i - 1], arr[i]};
            //call function for finding the octal in three
            int ans = binaryToOctal(temp, 3);
            //appending to new string  after converting the integer to char
            res += (ans + '0');
            i -= 2;
        }
        //similarly add two 0 then one value for size 1 left
        else if (i == 0)
        {
            char temp[3] = {0, 0, arr[i]};
            //call function for finding the octal in three
            int ans = binaryToOctal(temp, 3);
            //appending to new string  after converting the integer to char
            res += (ans + '0');
            i--;
        }
        else
        {
            i--;
        }
    }
    //reverse the resultant string to get the octal values
    reverse(res.begin(), res.end());
    cout << res << endl;
    return 0;
}

Output:

Thanks.