Given an octal number as input, we need to write a program to convert the given octal number into equivalent decimal number.
Examples:
Input : 67
Output: 55
Input : 512
Output: 330
Input : 123
Output: 83
The idea is to extract the digits of a given octal number starting from the rightmost digit and keep a variable dec_value. At the time of extracting digits from the octal number, multiply the digit with the proper base (Power of 8) and add it to the variable dec_value. In the end, the variable dec_value will store the required decimal number.
For Example:
If the octal number is 67.
dec_value = 6*(8^1) + 7*(8^0) = 55
The below diagram explains how to convert an octal number (123) to an equivalent decimal value:

Below is the implementation of the above idea.
C++
#include <iostream>
using namespace std;
int octalToDecimal( int n)
{
int num = n;
int dec_value = 0;
int base = 1;
int temp = num;
while (temp) {
int last_digit = temp % 10;
temp = temp / 10;
dec_value += last_digit * base;
base = base * 8;
}
return dec_value;
}
int main()
{
int num = 67;
cout << octalToDecimal(num) << endl;
}
|
Java
import java.io.*;
class GFG {
static int octalToDecimal( int n)
{
int num = n;
int dec_value = 0 ;
int base = 1 ;
int temp = num;
while (temp > 0 ) {
int last_digit = temp % 10 ;
temp = temp / 10 ;
dec_value += last_digit * base;
base = base * 8 ;
}
return dec_value;
}
public static void main(String[] args)
{
int num = 67 ;
System.out.println(octalToDecimal(num));
}
}
|
Python3
def octalToDecimal(n):
num = n
dec_value = 0
base = 1
temp = num
while (temp):
last_digit = temp % 10
temp = int (temp / 10 )
dec_value + = last_digit * base
base = base * 8
return dec_value
num = 67
print (octalToDecimal(num))
|
C#
using System;
class GFG {
static int octalToDecimal( int n)
{
int num = n;
int dec_value = 0;
int b_ase = 1;
int temp = num;
while (temp > 0) {
int last_digit = temp % 10;
temp = temp / 10;
dec_value += last_digit * b_ase;
b_ase = b_ase * 8;
}
return dec_value;
}
public static void Main()
{
int num = 67;
Console.WriteLine(octalToDecimal(num));
}
}
|
Javascript
<script>
function octalToDecimal(n)
{
let num = n;
let dec_value = 0;
let base = 1;
let temp = num;
while (temp) {
let last_digit = temp % 10;
temp = Math.floor(temp / 10);
dec_value += last_digit * base;
base = base * 8;
}
return dec_value;
}
let num = 67;
document.write(octalToDecimal(num) + "<br>" );
</script>
|
PHP
<?php
function octalToDecimal( $n )
{
$num = $n ;
$dec_value = 0;
$base = 1;
$temp = $num ;
while ( $temp )
{
$last_digit = $temp % 10;
$temp = $temp / 10;
$dec_value += $last_digit * $base ;
$base = $base * 8;
}
return $dec_value ;
}
$num = 67;
echo octalToDecimal( $num );
?>
|
Time complexity: O(logN) where N is the given number
Auxiliary space: O(1)
Method: Using look up table method
The function octalToDecimal takes an integer n as input, which represents the octal number that needs to be converted to decimal. It initializes an unordered map lookup that maps each octal digit to its decimal equivalent.
It then uses a loop to extract each digit of the octal number from right to left, starting from the least significant digit. For each digit, it multiplies the decimal equivalent of the digit (retrieved from the lookup table) with the appropriate power of 8 (base) and adds it to the decimal variable. The base variable is updated after each iteration by multiplying it with 8. The loop continues until all digits have been processed.
Finally, the decimal variable is returned as the output of the function.
In the main function, an octal number octal_num is initialized and passed as an argument to the octalToDecimal function. The resulting decimal value is printed to the console.
C++
#include <iostream>
#include <unordered_map>
using namespace std;
int octalToDecimal( int n) {
unordered_map< int , int > lookup{
{0, 0}, {1, 1}, {2, 2}, {3, 3},
{4, 4}, {5, 5}, {6, 6}, {7, 7}
};
int decimal = 0;
int base = 1;
while (n > 0) {
int last_digit = n % 10;
decimal += lookup[last_digit] * base;
n /= 10;
base *= 8;
}
return decimal;
}
int main() {
int octal_num = 67;
cout << octalToDecimal(octal_num) << endl;
return 0;
}
|
Java
import java.util.HashMap;
public class GFG {
public static int octalToDecimal( int n) {
HashMap<Integer, Integer> lookup = new HashMap<>();
lookup.put( 0 , 0 );
lookup.put( 1 , 1 );
lookup.put( 2 , 2 );
lookup.put( 3 , 3 );
lookup.put( 4 , 4 );
lookup.put( 5 , 5 );
lookup.put( 6 , 6 );
lookup.put( 7 , 7 );
int decimal = 0 ;
int base = 1 ;
while (n > 0 ) {
int lastDigit = n % 10 ;
decimal += lookup.get(lastDigit) * base;
n /= 10 ;
base *= 8 ;
}
return decimal;
}
public static void main(String[] args) {
int octalNum = 67 ;
System.out.println(octalToDecimal(octalNum));
}
}
|
Python3
def octal_to_decimal(n):
lookup = {
0 : 0 , 1 : 1 , 2 : 2 , 3 : 3 ,
4 : 4 , 5 : 5 , 6 : 6 , 7 : 7
}
decimal = 0
base = 1
while n > 0 :
last_digit = n % 10
decimal + = lookup[last_digit] * base
n / / = 10
base * = 8
return decimal
def main():
octal_num = 67
print (octal_to_decimal(octal_num))
if __name__ = = "__main__" :
main()
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
public static int OctalToDecimal( int n)
{
Dictionary< int , int > lookup = new Dictionary< int , int >()
{
{0, 0}, {1, 1}, {2, 2}, {3, 3},
{4, 4}, {5, 5}, {6, 6}, {7, 7}
};
int decimalNum = 0;
int baseVal = 1;
while (n > 0)
{
int lastDigit = n % 10;
decimalNum += lookup[lastDigit] * baseVal;
n /= 10;
baseVal *= 8;
}
return decimalNum;
}
public static void Main( string [] args)
{
int octalNum = 67;
int decimalValue = OctalToDecimal(octalNum);
Console.WriteLine( "Equivalent Decimal Value = " + decimalValue);
}
}
|
Javascript
function octalToDecimal(n) {
let lookup = new Map([
[0, 0], [1, 1], [2, 2], [3, 3],
[4, 4], [5, 5], [6, 6], [7, 7]
]);
let decimal = 0;
let base = 1;
while (n > 0) {
let last_digit = n % 10;
decimal += lookup.get(last_digit) * base;
n = Math.floor(n/10);
base *= 8;
}
return decimal;
}
let octal_num = 67;
document.write(octalToDecimal(octal_num));
|
Time complexity: The time complexity of this algorithm is O(log N), where N is the octal number being converted to decimal. This is because we loop through each digit in the octal number once, and the number of digits in an N-digit octal number is log N.
Auxiliary space: The space complexity of this algorithm is O(1), as we only store a fixed-size lookup table and a few integer variables for the running sum and base value.
Using predefined function
C++
#include <iostream>
using namespace std;
int OctToDec(string n)
{
return stoi(n, 0, 8);
}
int main()
{
string n = "67" ;
cout << OctToDec(n);
return 0;
}
|
Java
import java.io.*;
class GFG {
public static int OctToDec(String n)
{
return Integer.parseInt(n, 8 );
}
public static void main(String[] args)
{
String n = "67" ;
System.out.println(OctToDec(n));
}
}
|
Python3
def OctToDec(n):
return int (n, 8 );
if __name__ = = '__main__' :
n = "67" ;
print (OctToDec(n));
|
C#
using System;
public class GFG{
public static int OctToDec(String n)
{
return Convert.ToInt32(n, 8);
}
static public void Main (){
string n = "67" ;
Console.WriteLine(OctToDec(n));
}
}
|
Javascript
<script>
function OctToDec(n)
{
return parseInt(n, 8);
}
var n = "67" ;
document.write(OctToDec(n));
</script>
|
Method 3: Using recursion
1. The method uses the fact that each digit of an octal number represents a power of 8, starting from the rightmost digit.
2. The method extracts the rightmost digit of the octal number by taking the remainder of the number divided by 10 (i.e., octal % 10) and adds it to the product of the remaining digits and the appropriate power of 8 (i.e., 8 * octal_to_decimal(octal // 10)).
3.This recursive step continues until the entire number has been converted to decimal.
C++
#include <iostream>
using namespace std;
int octal_to_decimal( int octal) {
if (octal == 0) {
return 0;
} else {
return (octal % 10) + 8 * octal_to_decimal(octal / 10);
}
}
int main() {
int octal = 67;
cout << octal_to_decimal(octal) << endl;
return 0;
}
|
Java
public class OctalToDecimal {
public static int octalToDecimal( int octal) {
if (octal == 0 ) {
return 0 ;
} else {
return (octal % 10 ) + 8 * octalToDecimal(octal / 10 );
}
}
public static void main(String[] args) {
int octal = 67 ;
System.out.println(octalToDecimal(octal));
}
}
|
Python3
def octal_to_decimal(octal):
if octal = = 0 :
return 0
else :
return (octal % 10 ) + 8 * octal_to_decimal(octal / / 10 )
octal = 67
print (octal_to_decimal(octal))
|
C#
using System;
public class Program
{
public static void Main()
{
int octal = 67;
int decimalNum = OctalToDecimal(octal);
Console.WriteLine(decimalNum);
}
public static int OctalToDecimal( int octal)
{
if (octal == 0)
{
return 0;
}
else
{
return (octal % 10) + 8 * OctalToDecimal(octal / 10);
}
}
}
|
Javascript
function octal_to_decimal(octal) {
if (octal == 0) {
return 0;
} else {
return (octal % 10) + 8 * octal_to_decimal(Math.floor(octal / 10));
}
}
let octal = 67;
console.log(octal_to_decimal(octal));
|
The time complexity of this method is O(log n), where n is the given number
Auxiliary space: O(log n), where n is the given number
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