Related Articles

# Recursive Implementation of atoi()

• Difficulty Level : Basic
• Last Updated : 06 Nov, 2020

The atoi() function takes a string (which represents an integer) as an argument and returns its value.

We have discussed iterative implementation of atoi(). How to compute recursively?

We strongly recommend you to minimize your browser and try this yourself first
The idea is to separate the last digit, recursively compute the result for remaining n-1 digits, multiply the result with 10 and add the obtained value to the last digit.

Below is the implementation of the idea.

## C++

 `// Recursive C program to compute atoi()``#include ``#include `` ` `// Recursive function to compute atoi()``int` `myAtoiRecursive(``char` `*str, ``int` `n)``{``    ``// Base case (Only one digit)``    ``if` `(n == 1)``        ``return` `*str - ``'0'``;`` ` `    ``// If more than 1 digits, recur for (n-1), multiplu result with 10``    ``// and add last digit``    ``return` `(10 * myAtoiRecursive(str, n - 1) + str[n-1] - ``'0'``);``}`` ` `// Driver Program``int` `main(``void``)``{``    ``char` `str[] = ``"112"``;``    ``int` `n = ``strlen``(str);``    ``printf``(``"%d"``, myAtoiRecursive(str, n));``    ``return` `0;``}`

## Java

 `// Recursive Java program to compute atoi()``class` `GFG{`` ` `// Recursive function to compute atoi()``static` `int` `myAtoiRecursive(String str, ``int` `n)``{``     ` `    ``// Base case (Only one digit)``    ``if` `(n == ``1``)``    ``{``        ``return` `str.charAt(``0``) - ``'0'``;``    ``}``     ` `    ``// If more than 1 digits, recur for (n-1), ``    ``// multiplu result with 10 and add last digit``    ``return` `(``10` `* myAtoiRecursive(str, n - ``1``) + ``                      ``str.charAt(n - ``1``) - ``'0'``);``}`` ` `// Driver code``public` `static` `void` `main(String[] s)``{``    ``String str = ``"112"``;``    ``int` `n = str.length();``     ` `    ``System.out.println(myAtoiRecursive(str, n));``}``}`` ` `// This code is contributed by rutvik_56`

## Python3

 `# Python3 program to compute atoi()`` ` `# Recursive function to compute atoi()``def` `myAtoiRecursive(string, num):``     ` `    ``# base case, we've hit the end of the string,``    ``# we just return the last value``    ``if` `len``(string) ``=``=` `1``:``        ``return` `int``(string) ``+` `(num ``*` `10``)``         ` `    ``# add the next string item into our num value``    ``num ``=` `int``(string[``0``:``1``]) ``+` `(num ``*` `10``)``     ` `    ``# recurse through the rest of the string``    ``# and add each letter to num``    ``return` `myAtoiRecursive(string[``1``:], num)`` ` `# Driver Code    ``string ``=` `"112"`` ` `print``(myAtoiRecursive(string, ``0``))`` ` `# This code is contributed by Frank-Hu-MSFT`

## C#

 `// Recursive C# program to compute atoi()``using` `System;``class` `GFG{`` ` `// Recursive function to compute atoi()``static` `int` `myAtoiRecursive(``string` `str, ``int` `n)``{``     ` `    ``// Base case (Only one digit)``    ``if` `(n == 1)``    ``{``        ``return` `str - ``'0'``;``    ``}``     ` `    ``// If more than 1 digits, recur for (n-1), ``    ``// multiplu result with 10 and add last digit``    ``return` `(10 * myAtoiRecursive(str, n - 1) + ``                                  ``str[n - 1] - ``'0'``);``}`` ` `// Driver code``public` `static` `void` `Main()``{``    ``string` `str = ``"112"``;``    ``int` `n = str.Length;``     ` `    ``Console.Write(myAtoiRecursive(str, n));``}``}`` ` `// This code is contributed by Nidhi_Biet`

Output:

`112`