Write your own atoi() - GeeksforGeeks

The atoi() function in C takes a string (which represents an integer) as an argument and returns its value of type int. So basically the function is used to convert a string argument to an integer.

Syntax:

int atoi(const char strn)

Parameters: The function accepts one parameter strn which refers to the string argument that is needed to be converted into its integer equivalent.

Return Value: If strn is a valid input, then the function returns the equivalent integer number for the passed string number. If no valid conversion takes place, then the function returns zero.

Example:

#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
  
int main() 
{ 
    int val; 
    char strn1[] = "12546"; 
  
    val = atoi(strn1); 
    printf("String value = %s\n", strn1); 
    printf("Integer value = %d\n", val); 
  
    char strn2[] = "GeeksforGeeks"; 
    val = atoi(strn2); 
    printf("String value = %s\n", strn2); 
    printf("Integer value = %d\n", val); 
  
    return (0); 
} 

Output:

String value = 12546
Integer value = 12546
String value = GeeksforGeeks
Integer value = 0

Now let’s understand various ways in which one can create there own atoi() function supported by various conditions:

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Following is a simple implementation of conversion without considering any special case. We initialize result as 0. We start from the first character and update result for every character.

C

// Program to implement atoi() in C 
#include <stdio.h> 
  
// A simple atoi() function 
int myAtoi(char* str) 
{ 
    int res = 0; // Initialize result 
  
    // Iterate through all characters of input string and 
    // update result 
    for (int i = 0; str[i] != '\0'; ++i) 
        res = res * 10 + str[i] - '0'; 
  
    // return result. 
    return res; 
} 
  
// Driver program to test above function 
int main() 
{ 
    char str[] = "89789"; 
    int val = myAtoi(str); 
    printf("%d ", val); 
    return 0; 
} 

C++

// A simple C++ program for implementation of atoi 
#include <bits/stdc++.h> 
using namespace std; 
  
// A simple atoi() function 
int myAtoi(char* str) 
{ 
    int res = 0; // Initialize result 
  
    // Iterate through all characters of input string and 
    // update result 
    for (int i = 0; str[i] != '\0'; ++i) 
        res = res * 10 + str[i] - '0'; 
  
    // return result. 
    return res; 
} 
  
// Driver code 
int main() 
{ 
    char str[] = "89789"; 
    int val = myAtoi(str); 
    cout << val; 
    return 0; 
} 
  
// This is code is contributed by rathbhupendra 

Python

# Python program for implementation of atoi 
  
# A simple atoi() function 
def myAtoi(string): 
    res = 0
  
    # Iterate through all characters of input string and  
    # update result 
    for i in xrange(len(string)): 
        res = res * 10 + (ord(string[i]) - ord('0')) 
  
    return res 
  
# Driver program 
string = "89789"
print myAtoi(string) 
  
# This code is contributed by BHAVYA JAIN 

C#

// A simple C# program for implementation 
// of atoi 
using System; 
  
class GFG { 
  
    // A simple atoi() function 
    static int myAtoi(string str) 
    { 
        int res = 0; // Initialize result 
  
        // Iterate through all characters of 
        // input string and update result 
        for (int i = 0; i < str.Length; ++i) 
            res = res * 10 + str[i] - '0'; 
  
        // return result. 
        return res; 
    } 
  
    // Driver program to test above function 
    public static void Main() 
    { 
        string str = "89789"; 
        int val = myAtoi(str); 
        Console.Write(val); 
    } 
} 
  
// This code is contributed by Sam007. 

Output:

In the below implementation negative numbers have been handled.

C++

// A C++ program for implementation of atoi 
#include <bits/stdc++.h> 
using namespace std; 
  
// A simple atoi() function 
int myAtoi(char* str) 
{ 
    int res = 0; // Initialize result 
    int sign = 1; // Initialize sign as positive 
    int i = 0; // Initialize index of first digit 
  
    // If number is negative, then update sign 
    if (str[0] == '-') { 
        sign = -1; 
        i++; // Also update index of first digit 
    } 
  
    // Iterate through all digits and update the result 
    for (; str[i] != '\0'; ++i) 
        res = res * 10 + str[i] - '0'; 
  
    // Return result with sign 
    return sign * res; 
} 
  
// Driver code 
int main() 
{ 
    char str[] = "-123"; 
    int val = myAtoi(str); 
    cout << val; 
    return 0; 
} 
  
// This is code is contributed by rathbhupendra 

C

// A C program for implementation of atoi 
#include <stdio.h> 
  
// A simple atoi() function 
int myAtoi(char* str) 
{ 
    int res = 0; // Initialize result 
    int sign = 1; // Initialize sign as positive 
    int i = 0; // Initialize index of first digit 
  
    // If number is negative, then update sign 
    if (str[0] == '-') { 
        sign = -1; 
        i++; // Also update index of first digit 
    } 
  
    // Iterate through all digits and update the result 
    for (; str[i] != '\0'; ++i) 
        res = res * 10 + str[i] - '0'; 
  
    // Return result with sign 
    return sign * res; 
} 
  
// Driver program to test above function 
int main() 
{ 
    char str[] = "-123"; 
    int val = myAtoi(str); 
    printf("%d ", val); 
    return 0; 
}

Python

# Python program for implementation of atoi 
  
# A simple atoi() function 
def myAtoi(string): 
    res = 0
    # initialize sign as positive 
    sign = 1
    i = 0
  
    # if number is negative then update sign 
    if string[0] == '-': 
        sign = -1
        i+= 1
  
    # Iterate through all characters of input string and update result 
    for j in xrange(i, len(string)): 
        res = res * 10 + (ord(string[j]) - ord('0')) 
  
    return sign * res 
  
# Driver program 
string = "-123"
print myAtoi(string) 
  
# This code is contributed by BHAVYA JAIN 

C#

// C# program for implementation of atoi 
using System; 
  
class GFG { 
  
    // A simple atoi() function 
    static int myAtoi(string str) 
    { 
  
        // Initialize result 
        int res = 0; 
  
        // Initialize sign as positive 
        int sign = 1; 
  
        // Initialize index of first digit 
        int i = 0; 
  
        // If number is negative, then 
        // update sign 
        if (str[0] == '-') { 
            sign = -1; 
  
            // Also update index of first 
            // digit 
            i++; 
        } 
  
        // Iterate through all digits 
        // and update the result 
        for (; i < str.Length; ++i) 
            res = res * 10 + str[i] - '0'; 
  
        // Return result with sign 
        return sign * res; 
    } 
  
    // Driver program to test above function 
    public static void Main() 
    { 
        string str = "-123"; 
        int val = myAtoi(str); 
        Console.Write(val); 
    } 
} 
  
// This code is contributed by Sam007. 

Output:

-123

This implementation handles various type of errors. What if str is NULL or str contains non-numeric characters. Following implementation handles errors.

C++

// A simple C++ program for implementation of atoi 
  
#include <stdio.h> 
  
// A utility function to check whether x is numeric 
bool isNumericChar(char x) 
{ 
    return (x >= '0' && x <= '9') ? true : false; 
} 
  
// A simple atoi() function. If the given string contains 
// any invalid character, then this function returns 0 
int myAtoi(char* str) 
{ 
    if (*str == '\0') 
        return 0; 
  
    int res = 0; // Initialize result 
    int sign = 1; // Initialize sign as positive 
    int i = 0; // Initialize index of first digit 
  
    // If number is negative, then update sign 
    if (str[0] == '-') { 
        sign = -1; 
        i++; // Also update index of first digit 
    } 
  
    // Iterate through all digits of input string and update result 
    for (; str[i] != '\0'; ++i) { 
        if (isNumericChar(str[i]) == false) 
            return 0; // You may add some lines to write error message 
        // to error stream 
        res = res * 10 + str[i] - '0'; 
    } 
  
    // Return result with sign 
    return sign * res; 
} 
  
// Driver program to test above function 
int main() 
{ 
    char str[] = "-134"; 
    int val = myAtoi(str); 
    printf("%d ", val); 
    return 0; 
} 

Python

# Python program for implementation of atoi 
  
# A utility function to check whether x is numeric 
def isNumericChar(x): 
    if (x >= '0' and x <= '9'): 
        return True
    return False
  
# A simple atoi() function. If the given string contains 
# any invalid character, then this function returns 0 
def myAtoi(string): 
    if len(string) == 0: 
        return 0
  
    res = 0
    # initialize sign as positive 
    sign = 1
    i = 0
  
    # if number is negative then update sign 
    if string[0] == '-': 
        sign = -1
        i+= 1
  
    # Iterate through all characters of input string and update result 
    for j in xrange(i, len(string)): 
        # You may add some lines to write error message to error stream 
        if isNumericChar(string[j] == False): 
            return 0
  
        res = res * 10 + (ord(string[j]) - ord('0')) 
  
    return sign * res 
  
# Driver program 
string = "-134"
print myAtoi(string) 
  
# This code is contributed by BHAVYA JAIN 

Time Complexity: O(n) where n is the number of characters in the input string.

We only need to handle four cases:

Discards all leading whitespaces
sign of the number
overflow
invalid input

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

// A simple C++ program for implementation of atoi 
#include <bits/stdc++.h> 
  
using namespace std; 
int myAtoi(const char* str) 
{ 
    int sign = 1, base = 0, i = 0; 
    // if whitespaces then ignore. 
    while (str[i] == ' ') { 
        i++; 
    } 
    // sign of number 
    if (str[i] == '-' || str[i] == '+') { 
        sign = 1 - 2 * (str[i++] == '-'); 
    } 
    // checking for valid input 
    while (str[i] >= '0' && str[i] <= '9') { 
        // handling overflow test case 
        if (base > INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - '0' > 7)) { 
            if (sign == 1) 
                return INT_MAX; 
            else
                return INT_MIN; 
        } 
        base = 10 * base + (str[i++] - '0'); 
    } 
    return base * sign; 
} 
  
int main() 
{ 
    char str[] = "  -123"; 
    int val = myAtoi(str); 
    printf("%d ", val); 
    return 0; 
} 
// This code is contributed by Yogesh shukla. 

Output :

-123

Time Complexity O(n), where n is length of string.

Recursive program for atoi().

Exercise
Write your won atof() that takes a string (which represents an floating point value) as an argument and returns its value as double.

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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