C Program To Write Your Own atoi()

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:

C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    int val;
    char strn1[] = "12546";

    val = atoi(strn1);
    printf("String value = %s
", strn1);
    printf("Integer value = %d
", val);

    char strn2[] = "GeeksforGeeks";
    val = atoi(strn2);
    printf("String value = %s
", strn2);
    printf("Integer value = %d
", val);

    return (0);
}

Output

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

Time Complexity: O(1)

Auxiliary Space: O(1)

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

Approach 1: Following is a simple implementation of conversion without considering any special case. 

• Initialize the result as 0.
• Start from the first character and update result for every character.
• For every character update the answer as result = result * 10 + (s[i] – ‘0’)

C

// Program to implement atoi() in C
#include <stdio.h>

// A simple atoi() function
int myAtoi(char* str)
{
    // Initialize result
    int res = 0;

    // Iterate through all characters
    // of input string and update result
    // take ASCII character of corresponding digit and
    // subtract the code from '0' to get numerical
    // value and multiply res by 10 to shuffle
    // digits left to update running total
    for (int i = 0; str[i] != ''; ++i)
        res = res * 10 + str[i] - '0';

    // return result.
    return res;
}

// Driver Code
int main()
{
    char str[] = "89789";

    // Function call
    int val = myAtoi(str);
    printf("%d ", val);
    return 0;
}

Output

89789

Time Complexity: O(N)

Auxiliary Space: O(1)

Approach 2: This implementation handles the negative numbers. If the first character is ‘-‘ then store the sign as negative and then convert the rest of the string to number using the previous approach while multiplying sign with it. 

C

// A C program for
// implementation of atoi
#include <stdio.h>

// A simple atoi() function
int myAtoi(char* 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 (; str[i] != ''; ++i)
        res = res * 10 + str[i] - '0';

    // Return result with sign
    return sign * res;
}

// Driver code
int main()
{
    char str[] = "-123";

    // Function call
    int val = myAtoi(str);
    printf("%d ", val);
    return 0;
}

Output

-123

Time Complexity: O(N)

Auxiliary Space: O(1)

Approach 3: This implementation handles various type of errors. If str is NULL or str contains non-numeric characters then return 0 as the number is not valid. 

Output

 -134

Approach 4: Four corner cases needs to be handled: 

• Discards all leading whitespaces
• Sign of the number
• Overflow
• Invalid input

To remove the leading whitespaces run a loop until a character of the digit is reached. If the number is greater than or equal to INT_MAX/10. Then return INT_MAX if the sign is positive and return INT_MIN if the sign is negative. The other cases are handled in previous approaches. 

Dry Run:

Below is the implementation of the above approach: 

C

// A simple C++ program for
// implementation of atoi
#include <stdio.h>
#include <limits.h>

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;
}


// Driver Code
int main()
{
    char str[] = "  -123";

    // Functional Code
    int val = myAtoi(str);
    printf("%d ", val);
    return 0;
}
// This code is contributed by Yogesh shukla.

Output

 -123

Complexity Analysis for all the above Approaches:

• Time Complexity: O(n). 
  Only one traversal of string is needed.
• Space Complexity: O(1). 
  As no extra space is required.

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.

Please refer complete article on Write your own atoi() for more details!