Program to convert a binary number to octal

Last Updated : 05 Dec, 2023

The problem is to convert the given binary number (represented as string) to its equivalent octal number. The input could be very large and may not fit even into unsigned long long int.

Examples:

Input : 110001110
Output : 616

Input  : 1111001010010100001.010110110011011
Output : 1712241.26633

The idea is to consider the binary input as a string of characters and then follow the steps:

Get length of substring to the left and right of the decimal point(‘.’) as left_len and right_len.
If left_len is not a multiple of 3 add min number of 0’s in the beginning to make length of left substring a multiple of 3.
If right_len is not a multiple of 3 add min number of 0’s in the end to make length of right substring a multiple of 3.
Now, from the left extract one by one substrings of length 3 and add its corresponding octal code to the result.
If in between a decimal(‘.’) is encountered then add it to the result.

C++

// C++ implementation to convert a binary number 
// to octal number 
#include <bits/stdc++.h> 
using namespace std; 
   
// function to create map between binary 
// number and its equivalent octal 
void createMap(unordered_map<string, char> *um) 
{ 
    (*um)["000"] = '0'; 
    (*um)["001"] = '1'; 
    (*um)["010"] = '2'; 
    (*um)["011"] = '3'; 
    (*um)["100"] = '4'; 
    (*um)["101"] = '5'; 
    (*um)["110"] = '6'; 
    (*um)["111"] = '7';     
} 
   
// Function to find octal equivalent of binary 
string convertBinToOct(string bin) 
{ 
    int l = bin.size(); 
    int t = bin.find_first_of('.'); 
       
    // length of string before '.' 
    int len_left = t != -1 ? t : l; 
       
    // add min 0's in the beginning to make 
    // left substring length divisible by 3  
    for (int i = 1; i <= (3 - len_left % 3) % 3; i++) 
        bin = '0' + bin; 
       
    // if decimal point exists     
    if (t != -1)     
    { 
        // length of string after '.' 
        int len_right = l - len_left - 1; 
           
        // add min 0's in the end to make right 
        // substring length divisible by 3  
        for (int i = 1; i <= (3 - len_right % 3) % 3; i++) 
            bin = bin + '0'; 
    } 
       
    // create map between binary and its 
    // equivalent octal code 
    unordered_map<string, char> bin_oct_map; 
    createMap(&bin_oct_map); 
       
    int i = 0; 
    string octal = ""; 
       
    while (1) 
    { 
        // one by one extract from left, substring 
        // of size 3 and add its octal code 
        octal += bin_oct_map[bin.substr(i, 3)]; 
        i += 3; 
        if (i == bin.size()) 
            break; 
               
        // if '.' is encountered add it to result 
        if (bin.at(i) == '.')     
        { 
            octal += '.'; 
            i++; 
        } 
    } 
       
    // required octal number 
    return octal;     
} 
   
// Driver program to test above 
int main() 
{ 
    string bin = "1111001010010100001.010110110011011"; 
    cout << "Octal number = "
         << convertBinToOct(bin); 
    return 0;      
}   

Java

// Java implementation to convert a  
// binary number to octal number  
import java.io.*; 
import java.util.*; 
  
class GFG{ 
  
// Function to create map between binary 
// number and its equivalent hexadecimal 
static void createMap(Map<String, Character> um) 
{ 
    um.put("000", '0'); 
    um.put("001", '1'); 
    um.put("010", '2'); 
    um.put("011", '3'); 
    um.put("100", '4'); 
    um.put("101", '5'); 
    um.put("110", '6'); 
    um.put("111", '7'); 
} 
  
// Function to find octal equivalent of binary 
static String convertBinToOct(String bin) 
{ 
    int l = bin.length(); 
    int t = bin.indexOf('.'); 
  
    // Length of string before '.' 
    int len_left = t != -1 ? t : l; 
  
    // Add min 0's in the beginning to make 
    // left substring length divisible by 3 
    for(int i = 1;  
            i <= (3 - len_left % 3) % 3;  
            i++) 
        bin = '0' + bin; 
  
    // If decimal point exists 
    if (t != -1) 
    { 
          
        // Length of string after '.' 
        int len_right = l - len_left - 1; 
  
        // add min 0's in the end to make right 
        // substring length divisible by 3 
        for(int i = 1; 
                i <= (3 - len_right % 3) % 3; 
                i++) 
            bin = bin + '0'; 
    } 
  
    // Create map between binary and its 
    // equivalent octal code 
    Map<String,  
        Character> bin_oct_map = new HashMap<String,  
                                             Character>(); 
    createMap(bin_oct_map); 
  
    int i = 0; 
    String octal = ""; 
  
    while (true) 
    { 
          
        // One by one extract from left, substring 
        // of size 3 and add its octal code 
        octal += bin_oct_map.get( 
            bin.substring(i, i + 3)); 
        i += 3; 
          
        if (i == bin.length()) 
            break; 
  
        // If '.' is encountered add it to result 
        if (bin.charAt(i) == '.') 
        { 
            octal += '.'; 
            i++; 
        } 
    } 
  
    // Required octal number 
    return octal; 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    String bin = "1111001010010100001.010110110011011"; 
    System.out.println("Octal number = " + 
                        convertBinToOct(bin)); 
} 
} 
  
// This code is contributed by jithin

Python3

# Python3 implementation to convert a binary number  
# to octal number  
  
# function to create map between binary  
# number and its equivalent octal  
def createMap(bin_oct_map): 
    bin_oct_map["000"] = '0'
    bin_oct_map["001"] = '1'
    bin_oct_map["010"] = '2'
    bin_oct_map["011"] = '3'
    bin_oct_map["100"] = '4'
    bin_oct_map["101"] = '5'
    bin_oct_map["110"] = '6'
    bin_oct_map["111"] = '7'
  
# Function to find octal equivalent of binary  
def convertBinToOct(bin):  
    l = len(bin) 
      
    # length of string before '.'  
    t = -1
    if '.' in bin: 
        t = bin.index('.')  
        len_left = t 
    else: 
        len_left = l  
      
    # add min 0's in the beginning to make  
    # left substring length divisible by 3  
    for i in range(1, (3 - len_left % 3) % 3 + 1): 
        bin = '0' + bin
      
    # if decimal point exists  
    if (t != -1):  
          
        # length of string after '.'  
        len_right = l - len_left - 1
          
        # add min 0's in the end to make right  
        # substring length divisible by 3  
        for i in range(1, (3 - len_right % 3) % 3 + 1): 
            bin = bin + '0'
      
    # create dictionary between binary and its  
    # equivalent octal code  
    bin_oct_map = {} 
    createMap(bin_oct_map) 
    i = 0
    octal = "" 
      
    while (True) : 
          
        # one by one extract from left, substring  
        # of size 3 and add its octal code  
        octal += bin_oct_map[bin[i:i + 3]]  
        i += 3
        if (i == len(bin)):  
            break
              
        # if '.' is encountered add it to result  
        if (bin[i] == '.'): 
            octal += '.'
            i += 1
              
    # required octal number  
    return octal 
  
# Driver Code 
bin = "1111001010010100001.010110110011011"
print("Octal number = ",  
       convertBinToOct(bin)) 
  
# This code is contributed  
# by Atul_kumar_Shrivastava 

C#

// C# implementation to convert a  
// binary number to octal number  
  
using System; 
using System.Collections.Generic; 
  
public class GFG{ 
  
// Function to create map between binary 
// number and its equivalent hexadecimal 
static void createMap(Dictionary<String, char> um) 
{ 
    um.Add("000", '0'); 
    um.Add("001", '1'); 
    um.Add("010", '2'); 
    um.Add("011", '3'); 
    um.Add("100", '4'); 
    um.Add("101", '5'); 
    um.Add("110", '6'); 
    um.Add("111", '7'); 
} 
  
// Function to find octal equivalent of binary 
static String convertBinToOct(String bin) 
{ 
    int l = bin.Length; 
    int t = bin.IndexOf('.'); 
    int i = 0; 
    // Length of string before '.' 
    int len_left = t != -1 ? t : l; 
  
    // Add min 0's in the beginning to make 
    // left substring length divisible by 3 
    for(i = 1;  
            i <= (3 - len_left % 3) % 3;  
            i++) 
        bin = '0' + bin; 
  
    // If decimal point exists 
    if (t != -1) 
    { 
          
        // Length of string after '.' 
        int len_right = l - len_left - 1; 
  
        // add min 0's in the end to make right 
        // substring length divisible by 3 
        for(i = 1; 
                i <= (3 - len_right % 3) % 3; 
                i++) 
            bin = bin + '0'; 
    } 
  
    // Create map between binary and its 
    // equivalent octal code 
    Dictionary<String,  
        char> bin_oct_map = new Dictionary<String,  
                                             char>(); 
    createMap(bin_oct_map); 
  
    i = 0; 
    String octal = ""; 
  
    while (true) 
    { 
          
        // One by one extract from left, substring 
        // of size 3 and add its octal code 
        octal += bin_oct_map[ 
            bin.Substring(i,  3)]; 
        i += 3; 
          
        if (i == bin.Length) 
            break; 
  
        // If '.' is encountered add it to result 
        if (bin[i] == '.') 
        { 
            octal += '.'; 
            i++; 
        } 
    } 
  
    // Required octal number 
    return octal; 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    String bin = "1111001010010100001.010110110011011"; 
    Console.WriteLine("Octal number = " + 
                        convertBinToOct(bin)); 
} 
} 
  
// This code is contributed by 29AjayKumar

Javascript

<script> 
  
// Javascript implementation to convert a 
// binary number to octal number 
  
// Function to create map between binary 
// number and its equivalent hexadecimal 
function createMap(um)  
{ 
    um.set("000", '0'); 
    um.set("001", '1'); 
    um.set("010", '2'); 
    um.set("011", '3'); 
    um.set("100", '4'); 
    um.set("101", '5'); 
    um.set("110", '6'); 
    um.set("111", '7'); 
} 
  
// Function to find octal equivalent of binary 
function convertBinToOct(bin) 
{ 
      
    let l = bin.length; 
    let t = bin.indexOf('.'); 
  
    // Length of string before '.' 
    let len_left = t != -1 ? t : l; 
  
    // Add min 0's in the beginning to make 
    // left substring length divisible by 3 
    for(let i = 1;  
            i <= (3 - len_left % 3) % 3;  
            i++) 
        bin = '0' + bin; 
  
    // If decimal point exists 
    if (t != -1) 
    { 
          
        // Length of string after '.' 
        let len_right = l - len_left - 1; 
  
        // add min 0's in the end to make right 
        // substring length divisible by 3 
        for(let i = 1;  
                i <= (3 - len_right % 3) % 3; 
                i++) 
            bin = bin + '0'; 
    } 
  
    // Create map between binary and its 
    // equivalent octal code 
    let bin_oct_map = new Map(); 
  
    createMap(bin_oct_map); 
  
    let i = 0; 
    let octal = ""; 
  
    while (true)  
    { 
          
        // One by one extract from left, substring 
        // of size 3 and add its octal code 
        octal += bin_oct_map.get(bin.substr(i, 3)); 
        i += 3; 
  
        if (i == bin.length) 
            break; 
  
        // If '.' is encountered add it to result 
        if (bin.charAt(i) == '.') 
        { 
            octal += '.'; 
            i++; 
        } 
    } 
  
    // Required octal number 
    return octal; 
} 
  
// Driver code 
let bin = "1111001010010100001.010110110011011"; 
document.write("Octal number = " +  
               convertBinToOct(bin)); 
  
// This code is contributed by gfgking 
  
</script>