Question

How do i make this program accept strings?


#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;
}

Answer 1

#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()
{
    // **** I have updated here
    string arr;
    cout << "Enter the binary string: ";
    cin >> arr;

    int n = arr.size();

    // ****

    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:-

If you have any doubt, feel free to ask in the comments.