Write your own atoi()

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

Following is a simple implementation. We initialize result as 0. We start from the first character and update result for every character.

// A simple C++ program for implementation of atoi

#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;

}

Output:

89789

The above function doesn’t handle negative numbers. Following is a simple extension to handle negative numbers.

// 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;

}

Output:

-123

The above implementation doesn’t handle errors. What if str is NULL or str contains non-numeric characters. Following implementation handles errors.

// 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 == NULL)

       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;

}

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

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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//c# 

public static int myAtoi(string s)
        {
            if (string.IsNullOrEmpty(s))
            {
                return 0;
            }

            int sign = 1;
            int i = 0;
            int r= 0;
            //if number is negative,then update sign.
            if (s[0]=='-')
            {
                sign = -1;
                i++; //update index of first digit.
            }

            for (; i < s.Length; i++)
            {
                if (!(s[i]>='0'&&s[i]<='9'))
                {
                    return 0;
                }
                r = r * 10 + s[i]-'0'; //'1'=49;'0'=48; 'A'=65;'a'=97
            }
            return sign * r;
        }