Given a binary number as input, we need to write a program to convert the given binary number into an equivalent decimal number.
Examples :
Input : 111
Output : 7
Input : 1010
Output : 10
Input: 100001
Output: 33
The idea is to extract the digits of a given binary number starting from the rightmost digit and keep a variable dec_value. At the time of extracting digits from the binary number, multiply the digit with the proper base (Power of 2) 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 binary number is 111.
dec_value = 1*(2^2) + 1*(2^1) + 1*(2^0) = 7
The below diagram explains how to convert ( 1010 ) to equivalent decimal value:

Below is the implementation of the above idea :
C++
#include <iostream>
using namespace std;
int binaryToDecimal( 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 * 2;
}
return dec_value;
}
int main()
{
int num = 10101001;
cout << binaryToDecimal(num) << endl;
}
|
Java
class GFG {
static int binaryToDecimal( 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 * 2 ;
}
return dec_value;
}
public static void main(String[] args)
{
int num = 10101001 ;
System.out.println(binaryToDecimal(num));
}
}
|
Python3
def binaryToDecimal(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 * 2 ;
return dec_value;
num = 10101001 ;
print (binaryToDecimal(num));
|
C#
class GFG {
public static int binaryToDecimal( int n)
{
int num = n;
int dec_value = 0;
int base1 = 1;
int temp = num;
while (temp > 0) {
int last_digit = temp % 10;
temp = temp / 10;
dec_value += last_digit * base1;
base1 = base1 * 2;
}
return dec_value;
}
public static void Main()
{
int num = 10101001;
System.Console.Write(binaryToDecimal(num));
}
}
|
PHP
<?php
function binaryToDecimal( $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 *2;
}
return $dec_value ;
}
$num = 10101001;
echo binaryToDecimal( $num ), "\n" ;
?>
|
Javascript
<script>
function binaryToDecimal(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 * 2;
}
return dec_value;
}
let num = 10101001;
document.write(binaryToDecimal(num) + "<br>" );
</script>
|
Time complexity : O( logn)
Auxiliary Space : O(1)
Note: The program works only with binary numbers in the range of integers. In case you want to work with long binary numbers like 20 bits or 30 bit, you can use a string variable to store the binary numbers.
Below is a similar program which uses string variable instead of integers to store binary value:
C++
#include <iostream>
#include <string>
using namespace std;
int binaryToDecimal(string n)
{
string num = n;
int dec_value = 0;
int base = 1;
int len = num.length();
for ( int i = len - 1; i >= 0; i--) {
if (num[i] == '1' )
dec_value += base;
base = base * 2;
}
return dec_value;
}
int main()
{
string num = "10101001" ;
cout << binaryToDecimal(num) << endl;
}
|
Java
import java.io.*;
class GFG {
static int binaryToDecimal(String n)
{
String num = n;
int dec_value = 0 ;
int base = 1 ;
int len = num.length();
for ( int i = len - 1 ; i >= 0 ; i--) {
if (num.charAt(i) == '1' )
dec_value += base;
base = base * 2 ;
}
return dec_value;
}
public static void main(String[] args)
{
String num = new String( "10101001" );
System.out.println(binaryToDecimal(num));
}
}
|
Python3
def binaryToDecimal(n):
num = n;
dec_value = 0 ;
base1 = 1 ;
len1 = len (num);
for i in range (len1 - 1 , - 1 , - 1 ):
if (num[i] = = '1' ):
dec_value + = base1;
base1 = base1 * 2 ;
return dec_value;
num = "10101001" ;
print (binaryToDecimal(num));
|
C#
using System;
class GFG {
static int binaryToDecimal(String n)
{
String num = n;
int dec_value = 0;
int base1 = 1;
int len = num.Length;
for ( int i = len - 1; i >= 0; i--) {
if (num[i] == '1' )
dec_value += base1;
base1 = base1 * 2;
}
return dec_value;
}
public static void Main()
{
String num = "10101001" ;
Console.WriteLine(binaryToDecimal(num));
}
}
|
PHP
<?php
function binaryToDecimal( $n )
{
$num = $n ;
$dec_value = 0;
$base = 1;
$len = strlen ( $num );
for ( $i = $len - 1; $i >= 0; $i --)
{
if ( $num [ $i ] == '1' )
$dec_value += $base ;
$base = $base * 2;
}
return $dec_value ;
}
$num = "10101001" ;
echo binaryToDecimal( $num ), "\n" ;
?>
|
Javascript
<script>
function binaryToDecimal(n)
{
let num = n;
let dec_value = 0;
let base = 1;
let len = num.length;
for (let i = len - 1; i >= 0; i--) {
if (num[i] == '1' )
dec_value += base;
base = base * 2;
}
return dec_value;
}
let num = "10101001" ;
document.write(binaryToDecimal(num) + "<br>" );
</script>
|
Time complexity : O(n) where n is the length of the string.
Auxiliary Space : O(1)
Here is another way to convert decimal to binary numbers which is more intuitive and faster. At every iteration we extract the last digit (or a character in case of string input), check if it is a 1, if it is then multiply by the power of 2(where the power depends on which bit we are currently on ) and add it to the decimal number( initially = 0 ) .If it is a 0, then we don’t need to add anything to our decimal value since that bit is not contributing, so just increase the power and move to the next bit.
If we left shift 1 by n times ( 1<<n ), we basically get 2^n. Hence we use left shift operator( << ) to find the powers of two since it is faster.
1) If the input is of int type :
C++
#include <iostream>
using namespace std;
int binaryToDecimal( int n)
{
int dec_num = 0 ;
int power = 0 ;
while (n>0){
if (n%10 == 1){
dec_num += (1<<power) ;
}
power++ ;
n = n / 10 ;
}
return dec_num ;
}
int main()
{
int num = 10101001;
cout << binaryToDecimal(num) << endl;
}
|
Java
public class BinaryToDecimal {
public static int binaryToDecimal( int n)
{
int dec_num = 0 ;
int power = 0 ;
while (n > 0 ) {
if (n % 10 == 1 ) {
dec_num += ( 1 << power);
}
power++;
n = n / 10 ;
}
return dec_num;
}
public static void main(String[] args)
{
int num = 10101001 ;
System.out.println(binaryToDecimal(num));
}
}
|
Python3
import math
def binaryToDecimal(n):
dec_num = 0
power = 0
while (n > 0 ):
if (n % 10 = = 1 ):
dec_num + = ( 1 << power)
power = power + 1
n = math.floor(n / 10 )
return dec_num
num = 10101001
print (binaryToDecimal(num))
|
C#
using System;
public class Program
{
public static int BinaryToDecimal( int n)
{
int dec_num = 0;
int power = 0;
while (n > 0)
{
if (n % 10 == 1)
{
dec_num += (1 << power);
}
power++;
n = n / 10;
}
return dec_num;
}
public static void Main()
{
int num = 10101001;
Console.WriteLine(BinaryToDecimal(num));
}
}
|
Javascript
function binaryToDecimal(n) {
let dec_num = 0;
let power = 0;
while (n > 0) {
if (n % 10 === 1) {
dec_num += (1 << power);
}
power++;
n = Math.floor(n / 10);
}
return dec_num;
}
let num = 10101001;
console.log(binaryToDecimal(num));
|
Time complexity : O(log n)
Space complexity : O(1)
2) If the input is of string type :
C++
#include <iostream>
#include <string>
using namespace std;
int binaryToDecimal(string str)
{
int dec_num = 0;
int power = 0 ;
int n = str.length() ;
for ( int i = n-1 ; i>=0 ; i--){
if (str[i] == '1' ){
dec_num += (1<<power) ;
}
power++ ;
}
return dec_num;
}
int main()
{
string num = "10101001" ;
cout << binaryToDecimal(num) << endl;
}
|
Java
import java.util.*;
public class BinaryToDecimal {
public static int binaryToDecimal(String str) {
int dec_num = 0 ;
int power = 0 ;
int n = str.length();
for ( int i = n - 1 ; i >= 0 ; i--) {
if (str.charAt(i) == '1' ) {
dec_num += ( int ) Math.pow( 2 , power);
}
power++;
}
return dec_num;
}
public static void main(String[] args) {
String num = "10101001" ;
System.out.println(binaryToDecimal(num));
}
}
|
Python3
def binaryToDecimal( str ):
dec_num = 0
power = 0
n = len ( str )
for i in range (n - 1 , - 1 , - 1 ):
if ( str [i] = = '1' ):
dec_num + = ( 2 * * power)
power + = 1
return dec_num
num = "10101001"
print (binaryToDecimal(num))
|
C#
using System;
class Program {
static int binaryToDecimal( string str)
{
int dec_num = 0;
int power = 0;
int n = str.Length;
for ( int i = n - 1; i >= 0; i--) {
if (str[i] == '1' ) {
dec_num += (1 << power);
}
power++;
}
return dec_num;
}
static void Main( string [] args)
{
string num = "10101001" ;
Console.WriteLine(binaryToDecimal(num));
}
}
|
Javascript
function binaryToDecimal(str)
{
let dec_num = 0;
let power = 0 ;
let n = str.length;
for (let i = n-1 ; i>=0 ; i--){
if (str[i] == '1' ){
dec_num += (1<<power) ;
}
power++ ;
}
return dec_num;
}
let num = "10101001" ;
console.log(binaryToDecimal(num));
|
Time complexity: O(k) [since at max, we will have to process 32 bits which will take 32 iterations.]
Auxiliary Space : O(1)
Using pre-defined function:
C++
#include <iostream>
using namespace std;
int main()
{
char binaryNumber[] = "1001" ;
cout << stoi(binaryNumber, 0, 2);
return 0;
}
|
C
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int main()
{
char binaryNumber[] = "1001" ;
int bin, dec = 0;
bin = atoi (binaryNumber);
for ( int i = 0; bin; i++, bin /= 10)
if (bin % 10)
dec += pow (2, i);
printf ( "%d" , dec);
return 0;
}
|
Java
public class GFG {
public static void main(String args[])
{
String binaryNumber = "1001" ;
System.out.println(Integer.parseInt(binaryNumber, 2 ));
}
}
|
Python3
n = input ()
s = int (n, 2 )
print (s)
|
C#
using System;
class GFG {
public static void Main()
{
int value = 1001;
Console.Write(Convert.ToInt32(value.ToString(), 2));
}
}
|
Javascript
<script>
var binaryNumber = "1001" ;
document.write(parseInt(binaryNumber, 2));
</script>
|
Time complexity: O(n) where n is the length of the given string.
Auxiliary Space: O(1)
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