C++ Program To Write Your Own atoi() (original) (raw)

Last Updated : 23 Jul, 2025

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 <bits/stdc++.h> using namespace std;

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

val = atoi(strn1);
cout <<"String value = " << strn1 << endl;
cout <<"Integer value = " << val << endl;

char strn2[] = "GeeksforGeeks";
val = atoi(strn2);
cout <<"String value = " << strn2 << endl;
cout <<"Integer value = " << val <<endl;

return (0);

}

// This code is contributed by shivanisinghss2110

Output

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

Time Complexity : O(N)

Auxiliary Space : O(1) , as no extra space is needed.

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++ `

// A simple C++ program for // implementation of atoi #include <bits/stdc++.h> using namespace std;

// 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);
cout << val;
return 0;

}

// This is code is contributed by rathbhupendra

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 <bits/stdc++.h> using namespace std;

// 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);
cout << val;
return 0;

}

// This is code is contributed by rathbhupendra

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.

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 <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;

}

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

// Functional Code
int val = myAtoi(str);
cout <<" "<< val;
return 0;

}

// This code is contributed by shivanisinghss2110

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!