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C Program To Write Your Own atoi()

  • Last Updated : 23 Dec, 2021

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

Now let’s understand various ways in which one can create there 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

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

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!


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