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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?

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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`

This article is contributed by Narendra Kangralkar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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