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;

        }