# Program to Convert BCD number into Decimal number

Given a BCD (Binary Coded Decimal) number, the task is to convert the BCD number into it’s equivalent Decimal number.

Examples:

Input: BCD = 100000101000
Output: 828
Explanation:
Dividing the number into chunks of 4, it becomes 1000 0010 1000.
Here, 1000 is equivalent to 8 and
0010 is equivalent to 2.
So, the number becomes 828.

Input: BCD = 1001000
Output: 48
Explanation:
Dividing the number into chunks of 4, it becomes 0100 1000.
Here, 0100 is equivalent to 4 and
1000 is equivalent to 8.
So, the number becomes 48.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Iterate over all bits in given BCD numbers.
2. Divide the given BCD number into chunks of 4, and start computing its equivalent Decimal number.
3. Store this number formed in a variable named sum.
4. Start framing a number from the digits stored in sum in a variable num.
5. Reverse the number formed so far and return that number.

Below is the implementation of the above approach.

## C++

 `// C++ code to convert BCD to its ` `// decimal number(base 10). ` ` `  `// Including Header Files ` `#include ` `using` `namespace` `std; ` ` `  `// Function to convert BCD to Decimal ` `int` `bcdToDecimal(string s) ` `{ ` `    ``int` `len = s.length(), ` `        ``check = 0, check0 = 0; ` `    ``int` `num = 0, sum = 0, ` `        ``mul = 1, rev = 0; ` ` `  `    ``// Iterating through the bits backwards ` `    ``for` `(``int` `i = len - 1; i >= 0; i--) { ` ` `  `        ``// Forming the equivalent ` `        ``// digit(0 to 9) ` `        ``// from the group of 4. ` `        ``sum += (s[i] - ``'0'``) * mul; ` `        ``mul *= 2; ` `        ``check++; ` ` `  `        ``// Reinitialize all varaibles ` `        ``// and compute the number. ` `        ``if` `(check == 4 || i == 0) { ` `            ``if` `(sum == 0 && check0 == 0) { ` `                ``num = 1; ` `                ``check0 = 1; ` `            ``} ` `            ``else` `{ ` `                ``// update the answer ` `                ``num = num * 10 + sum; ` `            ``} ` ` `  `            ``check = 0; ` `            ``sum = 0; ` `            ``mul = 1; ` `        ``} ` `    ``} ` ` `  `    ``// Reverse the number formed. ` `    ``while` `(num > 0) { ` `        ``rev = rev * 10 + (num % 10); ` `        ``num /= 10; ` `    ``} ` ` `  `    ``if` `(check0 == 1) ` `        ``return` `rev - 1; ` ` `  `    ``return` `rev; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s = ``"100000101000"``; ` ` `  `    ``// Function Call ` `    ``cout << bcdToDecimal(s); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java code to convert BCD to its ` `// decimal number(base 10). ` `// Including Header Files ` `import` `java.io.*;  ` `import` `java.util.*; ` ` `  `class` `GFG {  ` `     `  `// Function to convert BCD to Decimal ` `public` `static` `int` `bcdToDecimal(String s) ` `{ ` `    ``int` `len = s.length(); ` `    ``int` `check = ``0``, check0 = ``0``; ` `    ``int` `num = ``0``, sum = ``0``; ` `    ``int` `mul = ``1``, rev = ``0``; ` ` `  `    ``// Iterating through the bits backwards ` `    ``for``(``int` `i = len - ``1``; i >= ``0``; i--)  ` `    ``{ ` ` `  `       ``// Forming the equivalent ` `       ``// digit(0 to 9) ` `       ``// from the group of 4. ` `       ``sum += (s.charAt(i) - ``'0'``) * mul; ` `       ``mul *= ``2``; ` `       ``check++; ` `        `  `       ``// Reinitialize all varaibles ` `       ``// and compute the number. ` `       ``if` `(check == ``4` `|| i == ``0``)  ` `       ``{ ` `           ``if` `(sum == ``0` `&& check0 == ``0``)  ` `           ``{ ` `               ``num = ``1``; ` `               ``check0 = ``1``; ` `           ``} ` `           ``else` `           ``{ ` ` `  `               ``// Update the answer ` `               ``num = num * ``10` `+ sum; ` `           ``} ` `            `  `           ``check = ``0``; ` `           ``sum = ``0``; ` `           ``mul = ``1``; ` `       ``} ` `    ``} ` `     `  `    ``// Reverse the number formed. ` `    ``while` `(num > ``0``)  ` `    ``{ ` `        ``rev = rev * ``10` `+ (num % ``10``); ` `        ``num /= ``10``; ` `    ``} ` ` `  `    ``if` `(check0 == ``1``) ` `        ``return` `rev - ``1``; ` ` `  `    ``return` `rev; ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``String s = ``"100000101000"``; ` ` `  `    ``// Function Call  ` `    ``System.out.println(bcdToDecimal(s)); ` `}  ` `}  ` ` `  `// This code is contributed by coder001 `

## Python3

 `# Python3 code to convert BCD to its  ` `# decimal number(base 10).  ` ` `  `# Function to convert BCD to Decimal  ` `def` `bcdToDecimal(s):  ` ` `  `    ``length ``=` `len``(s); ` `    ``check ``=` `0``; ` `    ``check0 ``=` `0``; ` `    ``num ``=` `0``; ` `    ``sum` `=` `0``; ` `    ``mul ``=` `1``; ` `    ``rev ``=` `0``; ` `     `  `    ``# Iterating through the bits backwards ` `    ``for` `i ``in` `range``(length ``-` `1``, ``-``1``, ``-``1``): ` `         `  `        ``# Forming the equivalent  ` `        ``# digit(0 to 9) ` `        ``# from the group of 4. ` `        ``sum` `+``=` `(``ord``(s[i]) ``-` `ord``(``'0'``)) ``*` `mul; ` `        ``mul ``*``=` `2``; ` `        ``check ``+``=` `1``; ` `         `  `        ``# Reinitialize all varaibles ` `        ``# and compute the number ` `        ``if` `(check ``=``=` `4` `or` `i ``=``=` `0``): ` `            ``if` `(``sum` `=``=` `0` `and` `check0 ``=``=` `0``): ` `                ``num ``=` `1``; ` `                ``check0 ``=` `1``; ` `                 `  `            ``else``: ` `                 `  `                ``# Update the answer ` `                ``num ``=` `num ``*` `10` `+` `sum``; ` `                 `  `            ``check ``=` `0``; ` `            ``sum` `=` `0``; ` `            ``mul ``=` `1``; ` `             `  `    ``# Reverse the number formed. ` `    ``while` `(num > ``0``): ` `        ``rev ``=` `rev ``*` `10` `+` `(num ``%` `10``); ` `        ``num ``/``/``=` `10``; ` `         `  `    ``if` `(check0 ``=``=` `1``): ` `        ``return` `rev ``-` `1``; ` `         `  `    ``return` `rev;  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``s ``=` `"100000101000"``; ` `     `  `    ``# Function Call ` `    ``print``(bcdToDecimal(s)); ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# code to convert BCD to its ` `// decimal number(base 10). ` `// Including Header Files ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `// Function to convert BCD to Decimal ` `public` `static` `int` `bcdToDecimal(String s) ` `{ ` `    ``int` `len = s.Length; ` `    ``int` `check = 0, check0 = 0; ` `    ``int` `num = 0, sum = 0; ` `    ``int` `mul = 1, rev = 0; ` ` `  `    ``// Iterating through the bits backwards ` `    ``for``(``int` `i = len - 1; i >= 0; i--)  ` `    ``{ ` ` `  `        ``// Forming the equivalent ` `        ``// digit(0 to 9) ` `        ``// from the group of 4. ` `        ``sum += (s[i] - ``'0'``) * mul; ` `        ``mul *= 2; ` `        ``check++; ` `             `  `        ``// Reinitialize all varaibles ` `        ``// and compute the number. ` `        ``if` `(check == 4 || i == 0)  ` `        ``{ ` `            ``if` `(sum == 0 && check0 == 0)  ` `            ``{ ` `                ``num = 1; ` `                ``check0 = 1; ` `            ``} ` `            ``else` `            ``{ ` `     `  `                ``// Update the answer ` `                ``num = num * 10 + sum; ` `            ``} ` `             `  `            ``check = 0; ` `            ``sum = 0; ` `            ``mul = 1; ` `        ``} ` `    ``} ` `     `  `    ``// Reverse the number formed. ` `    ``while` `(num > 0)  ` `    ``{ ` `        ``rev = rev * 10 + (num % 10); ` `        ``num /= 10; ` `    ``} ` ` `  `    ``if` `(check0 == 1) ` `        ``return` `rev - 1; ` ` `  `    ``return` `rev; ` `} ` ` `  `// Driver code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``String s = ``"100000101000"``; ` ` `  `    ``// Function Call  ` `    ``Console.WriteLine(bcdToDecimal(s)); ` `}  ` `}  ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```828
```

Time complexity: O(N)
Auxiliary Space complexity: O(1)

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