# Position of rightmost bit with first carry in sum of two binary

Given two non-negative integers a and b. The problem is to find the position of the rightmost bit where a carry is generated in the binary addition of a and b.

Examples:

```Input : a = 10, b = 2
Output : 2
(10)10 = (1010)2
(2)10 = (10)2.
1010
+   10
As highlighted, 1st carry bit from the right
will be generated at position '2'.

Input : a = 10, b = 5
Output : 0
'0' as no carry bit will be generated.
```

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

Approach: Following are the steps:

1. Calculate num = a & b.
2. Find the position of rightmost set bit in num.

## CPP

 `// C++ implementation to find the position of ` `// rightmost bit where a carry is generated first ` `#include ` `using` `namespace` `std; ` ` `  `typedef` `unsigned ``long` `long` `int` `ull; ` ` `  `// function to find the position of ` `// rightmost set bit in 'n' ` `unsigned ``int` `posOfRightmostSetBit(ull n) ` `{ ` `    ``return` `log2(n & -n) + 1; ` `} ` ` `  `// function to find the position of rightmost ` `// bit where a carry is generated first ` `unsigned ``int` `posOfCarryBit(ull a, ull b) ` `{ ` `    ``return` `posOfRightmostSetBit(a & b); ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``ull a = 10, b = 2; ` `    ``cout << posOfCarryBit(a, b); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the position of ` `// rightmost bit where a carry is generated first ` `class` `GFG { ` `     `  `    ``// function to find the position of ` `    ``// rightmost set bit in 'n' ` `    ``static` `int` `posOfRightmostSetBit(``int` `n) ` `    ``{ ` `        ``return` `(``int``)(Math.log(n & -n) / Math.log(``2``)) + ``1``; ` `    ``} ` ` `  `    ``// function to find the position of rightmost ` `    ``// bit where a carry is generated first ` `    ``static` `int` `posOfCarryBit(``int` `a, ``int` `b) ` `    ``{ ` `        ``return` `posOfRightmostSetBit(a & b); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `a = ``10``, b = ``2``; ` `         `  `        ``System.out.print(posOfCarryBit(a, b)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python3 implementation to find the position of ` `# rightmost bit where a carry is generated first ` ` `  `import` `math ` ` `  `# function to find the position of ` `# rightmost set bit in 'n' ` `def` `posOfRightmostSetBit( n ): ` `    ``return` `int``(math.log2(n & ``-``n) ``+` `1``) ` `     `  `# function to find the position of rightmost ` `# bit where a carry is generated first ` `def` `posOfCarryBit( a , b ): ` `    ``return` `posOfRightmostSetBit(a & b) ` ` `  `# Driver program to test above ` `a ``=` `10` `b ``=` `2` `print``(posOfCarryBit(a, b)) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// C# implementation to find the position of ` `// rightmost bit where a carry is generated first ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// function to find the position of ` `    ``// rightmost set bit in 'n' ` `    ``static` `int` `posOfRightmostSetBit(``int` `n) ` `    ``{ ` `        ``return` `(``int``)(Math.Log(n & -n) / Math.Log(2)) + 1; ` `    ``} ` `     `  `    ``// function to find the position of rightmost ` `    ``// bit where a carry is generated first ` `    ``static` `int` `posOfCarryBit(``int` `a, ``int` `b) ` `    ``{ ` `        ``return` `posOfRightmostSetBit(a & b); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `a = 10, b = 2; ` `         `  `        ``Console.Write(posOfCarryBit(a, b)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007. `

Output:

```2
```

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