# Convert Binary fraction to Decimal

Given an string of binary number n. Convert binary fractional n into it’s decimal equivalent.

```Input: n = 110.101
Output: 6.625

Input: n = 101.1101
Output: 5.8125
```
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Following are the steps of converting binary fractional to decimal.

A) Convert the integral part of binary to decimal equivalent

1. Multiply each digit separately from left side of radix point till the first digit by 20, 21, 22,… respectively.
2. Add all the result coming from step 1.
3. Equivalent integral decimal number would be the result obtained in step 2.

B) Convert the fractional part of binary to decimal equivalent

1. Divide each digit from right side of radix point till the end by 21, 22, 23, … respectively.
2. Add all the result coming from step 1.
3. Equivalent fractional decimal number would be the result obtained in step 2.

C) Add both integral and fractional part of decimal number.

Illustration

```Let's take an example for n = 110.101

Step 1: Conversion of 110 to decimal
=> 1102 = (1*22) + (1*21) + (0*20)
=> 1102 = 4 + 2 + 0
=> 1102 = 6
So equivalent decimal of binary integral is 6.

Step 2: Conversion of .0101 to decimal
=> 0.1012 = (1*1/2) + (0*1/22) + (1*1/23)
=> 0.1012 = 1*0.5 + 0*0.25 + 1*0.125
=> 0.1012 = 0.625
So equivalent decimal of binary fractional is 0.625

Step 3: Add result of step 1 and 2.
=> 6 + 0.625 = 6.625
```
```// C++ program to demonstrate above steps of
// binary fractional to decimal conversion
#include<bits/stdc++.h>
using namespace std;

// Function to convert binary fractional to
// decimal
double binaryToDecimal(string binary, int len)
{
// Fetch the radix point
size_t point = binary.find('.');

if (point == string::npos)
point = len;

double intDecimal = 0, fracDecimal = 0, twos = 1;

// Convert integral part of binary to decimal
// equivalent
for (int i = point-1; i>=0; --i)
{
// Subtract '0' to convert character
// into integer
intDecimal += (binary[i] - '0') * twos;
twos *= 2;
}

// Convert fractional part of binary to
// decimal equivalent
twos = 2;
for (int i = point+1; i < len; ++i)
{
fracDecimal += (binary[i] - '0') / twos;
twos *= 2.0;
}

// Add both integral and fractional part
return intDecimal + fracDecimal;
}

// Driver code
int main()
{
string n = "110.101";
cout << binaryToDecimal(n, n.length()) << "\n";

n = "101.1101";
cout << binaryToDecimal(n, n.length());

return 0;
}
```
```Output:
6.625
5.8125
```

Time complexity: O(len(n))
Auxiliary space: O(len(n))

Where len is the total digits contain in binary number of n.

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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