# Recursive Implementation of atoi()

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`

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Improved By : fhu004, rutvik_56, nidhi_biet

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