# Program for Binary To Decimal Conversion

Given a binary number as input, we need to write a program to convert the given binary number into an equivalent decimal number.

Check:

Binary to Decimal Converter

Examples :Â

`Input : 111Output : 7Input : 1010Output : 10Input: 100001Output: 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:Â
If the binary number is 111.Â
dec_value = 1*(2^2) + 1*(2^1) + 1*(2^0) = 7

The below diagram explains how to convert ( 1010 ) to equivalent decimal value:Â
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Below is the implementation of the above idea :Â

## C++

 `// C++ program to convert binary to decimal``#include ``using` `namespace` `std;` `// Function to convert binary to decimal``int` `binaryToDecimal(``int` `n)``{``    ``int` `num = n;``    ``int` `dec_value = 0;` `    ``// Initializing base value to 1, i.e 2^0``    ``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;``}` `// Driver program to test above function``int` `main()``{``    ``int` `num = 10101001;` `    ``cout << binaryToDecimal(num) << endl;``}`

## Java

 `// Java program to convert``// binary to decimal` `// Function to convert``// binary to decimal``class` `GFG {``    ``static` `int` `binaryToDecimal(``int` `n)``    ``{``        ``int` `num = n;``        ``int` `dec_value = ``0``;` `        ``// Initializing base``        ``// value to 1, i.e 2^0``        ``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;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `num = ``10101001``;``        ``System.out.println(binaryToDecimal(num));``    ``}``}` `// This code is contributed by mits.`

## Python3

 `# Python3 program to convert ``# binary to decimal` `# Function to convert ``# binary to decimal``def` `binaryToDecimal(n):``    ``num ``=` `n;``    ``dec_value ``=` `0``;``    ` `    ``# Initializing base ``    ``# value to 1, i.e 2 ^ 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;` `# Driver Code``num ``=` `10101001``;``print``(binaryToDecimal(num));` `# This code is contributed by mits`

## C#

 `// C# program to convert``// binary to decimal` `// Function to convert``// binary to decimal``class` `GFG {``    ``public` `static` `int` `binaryToDecimal(``int` `n)``    ``{``        ``int` `num = n;``        ``int` `dec_value = 0;` `        ``// Initializing base1``        ``// value to 1, i.e 2^0``        ``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;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `num = 10101001;` `        ``System.Console.Write(binaryToDecimal(num));``    ``}``}` `// This code is contributed by mits.`

## Javascript

 ``

## PHP

 ``

Output
```169
```

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:Â
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## C++

 `// C++ program to convert binary to decimal``// when input is represented as binary string.``#include ``#include ``using` `namespace` `std;` `// Function to convert binary to decimal``int` `binaryToDecimal(string n)``{``    ``string num = n;``    ``int` `dec_value = 0;` `    ``// Initializing base value to 1, i.e 2^0``    ``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;``}` `// Driver program to test above function``int` `main()``{``    ``string num = ``"10101001"``;``    ``cout << binaryToDecimal(num) << endl;``}`

## Java

 `// Java program to convert binary to``// decimal when input is represented``// as binary string.``import` `java.io.*;` `class` `GFG {` `    ``// Function to convert binary to decimal``    ``static` `int` `binaryToDecimal(String n)``    ``{``        ``String num = n;``        ``int` `dec_value = ``0``;` `        ``// Initializing base value to 1,``        ``// i.e 2^0``        ``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;``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String num = ``new` `String(``"10101001"``);``        ``System.out.println(binaryToDecimal(num));``    ``}``}` `// This code is contributed by Prerna Saini`

## Python3

 `# Python3 program to convert binary ``# to decimal when input is ``# represented as binary string.` `# Function to convert ``# binary to decimal``def` `binaryToDecimal(n):``    ``num ``=` `n;``    ``dec_value ``=` `0``;``    ` `    ``# Initializing base ``    ``# value to 1, i.e 2 ^ 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;` `# Driver Code``num ``=` `"10101001"``; ``print``(binaryToDecimal(num));` `# This code is contributed by mits`

## C#

 `// C# program to convert binary to``// decimal when input is represented``// as binary string.``using` `System;` `class` `GFG {` `    ``// Function to convert binary to decimal``    ``static` `int` `binaryToDecimal(String n)``    ``{``        ``String num = n;``        ``int` `dec_value = 0;` `        ``// Initializing base value to 1,``        ``// i.e 2^0``        ``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;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``String num = ``"10101001"``;``        ``Console.WriteLine(binaryToDecimal(num));``    ``}``}` `// This code is contribute by mits`

## Javascript

 ``

## PHP

 `= 0; ``\$i``--)``    ``{``        ``if` `(``\$num``[``\$i``] == ``'1'``)     ``            ``\$dec_value` `+= ``\$base``;``        ``\$base` `= ``\$base` `* 2;``    ``}``    ` `    ``return` `\$dec_value``;``}` `// Driver Code``\$num` `= ``"10101001"``; ``echo` `binaryToDecimal(``\$num``), ``"\n"``;` `// This code is contributed by aj_36``?>`

Output
```169
```

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++

 `// C++ program to convert binary to decimal``#include ``using` `namespace` `std;` `// Function to convert binary to decimal``int` `binaryToDecimal(``int` `n)``{``    ``int` `dec_num = 0 ;``      ``int` `power = 0 ;``      ``while``(n>0){``      ``if``(n%10 == 1){ ``// extracting the last digit``        ``dec_num += (1<

## 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));``  ``}``}` `// This code is contributed by aadityamaharshi21.`

## Python3

 `# Python program to convert binary to decimal``import` `math ` `def` `binaryToDecimal(n):``    ``dec_num ``=` `0``    ``power ``=` `0``    ``while` `(n > ``0``):``        ``if` `(n ``%` `10` `=``=` `1``): ``# extracting the last digit``            ``dec_num ``+``=` `(``1` `<< power)``        ``power ``=` `power ``+` `1``        ``n ``=` `math.floor(n ``/` `10``)``        ` `    ``return` `dec_num` `# Driver program to test the function``num ``=` `10101001` `print``(binaryToDecimal(num))` `# This code is contributed by Nidhi goel.`

## C#

 `using` `System;` `public` `class` `Program``{``  ``// Function to convert binary to decimal``  ``public` `static` `int` `BinaryToDecimal(``int` `n)``  ``{``    ``int` `dec_num = 0;``    ``int` `power = 0;` `    ``while` `(n > 0)``    ``{``      ``if` `(n % 10 == 1) ``// extracting the last digit``      ``{``        ``dec_num += (1 << power);``      ``}``      ``power++;``      ``n = n / 10;``    ``}` `    ``return` `dec_num;``  ``}` `  ``// Driver program to test above function``  ``public` `static` `void` `Main()``  ``{``    ``int` `num = 10101001;` `    ``Console.WriteLine(BinaryToDecimal(num));``  ``}``}`

## Javascript

 `// JavaScript program to convert binary to decimal``function` `binaryToDecimal(n) {``    ``let dec_num = 0;``    ``let power = 0;``    ``while` `(n > 0) {``        ``if` `(n % 10 === 1) { ``// extracting the last digit``            ``dec_num += (1 << power);``        ``}``        ``power++;``        ``n = Math.floor(n / 10);``    ``}``    ``return` `dec_num;``}` `// Driver program to test the function``let num = 10101001;` `console.log(binaryToDecimal(num));` `// This code is contributed by adityamaharshi21.`

Output
```169

```

Time complexity : O(log n)

Space complexity : O(1)

2) If the input is of string type :

## C++

 `// C++ program to convert binary to decimal``// when input is represented as binary string.``#include ``#include ``using` `namespace` `std;` `// Function to convert binary to decimal``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<

## Java

 `// Java program to convert binary to decimal``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

 `# Python program to convert binary to decimal ``def` `binaryToDecimal(``str``): ``    ``dec_num ``=` `0``    ``power ``=` `0``    ``n ``=` `len``(``str``) ` `    ``# Iterate through the string str ``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``): ``        ``if` `(``str``[i] ``=``=` `'1'``): ``            ``dec_num ``+``=` `(``2` `*``*` `power) ``        ``power ``+``=` `1``  ` `    ``return` `dec_num ``  ` `# Driver code ``num ``=` `"10101001"``print``(binaryToDecimal(num))`

## C#

 `using` `System;` `class` `Program {``    ``// Function to convert binary to decimal``    ``static` `int` `binaryToDecimal(``string` `str)``    ``{``        ``int` `dec_num = 0;``        ``int` `power = 0;``        ``int` `n = str.Length;` `        ``// Loop through the string from right to left``        ``for` `(``int` `i = n - 1; i >= 0; i--) {``            ``// If the current character is '1', add 2^power``            ``// to the decimal number``            ``if` `(str[i] == ``'1'``) {``                ``dec_num += (1 << power);``            ``}``            ``power++;``        ``}``        ``return` `dec_num;``    ``}` `    ``// Driver program to test above function``    ``static` `void` `Main(``string``[] args)``    ``{``        ``string` `num = ``"10101001"``;``        ``Console.WriteLine(binaryToDecimal(num));``    ``}``}`

## Javascript

 `// Javascript program to convert binary to decimal``// when input is represented as binary string.`  `// Function to convert binary to decimal``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<

Output
```169
```

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:Â
Â

## C++

 `#include ``using` `namespace` `std;` `int` `main()``{``    ``char` `binaryNumber[] = ``"1001"``; ``    ` `    ``cout << stoi(binaryNumber, 0, 2);` `    ``return` `0;``}``// This code is contributed by whysodarkbro`

## C

 `#include ``#include ``#include ` `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;``}` `// This code is contributed by whysodarkbro`

## Java

 `public` `class` `GFG {``    ``public` `static` `void` `main(String args[])``    ``{``        ``String binaryNumber = ``"1001"``;``        ``System.out.println(Integer.parseInt(binaryNumber, ``2``));``    ``}``}` `// This code is contributed``// By Yash Kumar Arora`

## Python3

 `n ``=` `input``()` `# Convert n to base 2``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));``} ``}` `// This code is contributed by SoumikMondal`

## Javascript

 ``

Output
`9`

Time complexity: O(n) where n is the length of the given string.

Auxiliary Space: O(1)Â

#### 3) Using Horner’s scheme

Hornerâ€™s scheme is an efficient approach for computing polynomials and it can be redirected into different other scenarios like converting numbers from one numerical system to another (Binary to decimal, etc.), evaluating the Taylor series, etc.

Given a polynomial of degree n, Cnxn + Cn-1xn-1 + Cn-2 xn-2 +â€¦+C1x + C0, it can be nested such that the solution is achieved with n multiplications and n additions using this scheme. Simply put, starting from the first coefficient, multiply by x and add the next coefficient.

`Cnxn + Cn-1xn-1 + Cn-2 xn-2 +â€¦+C1x + C0 = (((Cnx + Cn-1)x + Cn-2)x + Cn-3)x +â€¦ + C0`

Example:

3x3 + 4x2 + 6x + 3 = ((3x + 4)x + 6)x + 3

Observe that when we expand a binary number, it takes the form of a polynomial:

`Polynomial: Cnxn + Cn-1xn-1 + Cn-2xn-2 +â€¦+C1x + C0Binary expansion: digitn(2n) + digitn-1(2n-1) + digitn-2(2n-2) + â€¦ + digit0(20)`

The constants in the polynomial form are replaced by digits of the binary number and x=2. Hence, we can just use the same polynomial trick on this conversion:

Example:

101002

= 1*24 + 0*23 + 1*22 + 0*21 + 0*20

= (((1(2) + 0)2 + 1)2 + 0)2 + 0

= 20

## C++

 `#include ``#include ` `int` `main() {``    ``// Create a string to store the binary number``    ``std::string binaryNumber;` `    ``// Prompt the user to enter a binary number``    ``std::cout << ``"Enter Binary number: "``;` `    ``// Read the entire line as input``    ``std::getline(std::cin, binaryNumber);` `    ``// Convert binary to decimal``    ``int` `decimalNumber = 0;``    ``for` `(``size_t` `i = 0; i < binaryNumber.length(); i++) {``        ``decimalNumber = decimalNumber * 2 + (binaryNumber[i] - ``'0'``);``    ``}` `    ``// Print the decimal number``    ``std::cout << ``"Decimal: "` `<< decimalNumber << std::endl;` `    ``return` `0;``}`

## Java

 `import` `java.util.Scanner;` `public` `class` `BinaryToDecimal {``    ``public` `static` `void` `main(String[] args) {``        ``// Create a Scanner object to take user input``        ``Scanner scanner = ``new` `Scanner(System.in);` `        ``// Prompt the user to enter a binary number``        ``System.out.print(``"Enter Binary number: "``);``        ` `        ``// Read the entire line as input``        ``String binaryNumber = scanner.nextLine();` `        ``// Convert binary to decimal``        ``int` `decimalNumber = ``1``;``        ``for` `(``int` `i = ``0``; i < binaryNumber.length() - ``1``; i++) {``            ``decimalNumber = decimalNumber * ``2` `+ Character.getNumericValue(binaryNumber.charAt(i + ``1``));``        ``}` `        ``// Print the decimal number``        ``System.out.print(``"Decimal: "``);``        ``System.out.println(decimalNumber);` `        ``// Close the Scanner``        ``scanner.close();``    ``}``}`

## Python3

 `n ``=` `input``(``"Enter Binary number: "``)``dec ``=` `1``for` `i ``in` `range``(``len``(n)``-``1``):``    ``dec ``=` `dec``*``2` `+` `int``(n[i``+``1``])``print``(``"Decimal: "``, end ``=` `"")``print``(dec)` `# This code is contributed by Jeevani Anumandla`

## C#

 `using` `System;` `class` `Program {``    ``static` `void` `Main(``string``[] args)``    ``{``        ``// Create a string to store the binary number``        ``string` `binaryNumber;` `        ``// Prompt the user to enter a binary number``        ``Console.Write(``"Enter Binary number: "``);` `        ``// Read the entire line as input``        ``binaryNumber = Console.ReadLine();` `        ``// Check if input is empty``        ``if` `(``string``.IsNullOrEmpty(binaryNumber)) {``            ``Console.WriteLine(``"Decimal: 0"``);``            ``return``; ``// Exit the program``        ``}` `        ``// Convert binary to decimal``        ``int` `decimalNumber = 0;``        ``for` `(``int` `i = 0; i < binaryNumber.Length; i++) {``            ``decimalNumber = decimalNumber * 2``                            ``+ (binaryNumber[i] - ``'0'``);``        ``}` `        ``// Print the decimal number``        ``Console.WriteLine(``"Decimal: "` `+ decimalNumber);``    ``}``}`

## Javascript

 `// Prompt the user to enter a binary number``var` `n = prompt(``"Enter Binary number: "``);` `// Initialize the decimal variable to store the result``var` `dec = 1;` `// Iterate through each digit of the binary number``for` `(``var` `i = 0; i < n.length - 1; i++) {``// Update the decimal value using the formula: dec = dec * 2 + currentBinaryDigit``dec = dec * 2 + parseInt(n[i + 1]);``}` `// Display the decimal result``console.log(``"Decimal: "` `+ dec);`

Output:

Enter Binary number: 10010

Decimal: 18

Time complexity: O(n)

Decreased operations: Horners scheme only takes n multiplications and n additions for the conversion.

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