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Recursive Implementation of atoi()

  • Difficulty Level : Easy
  • 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 <stdio.h>
#include <string.h>
  
// 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] - '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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