C++ Program To Write Your Own atoi() - GeeksforGeeks

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 <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
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++
// A simple C++ program for// implementation of atoi#include <bits/stdc++.h>using namespace std;  // A simple atoi() functionint 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 codeint main(){    char str[] = "89789";        // Function call    int val = myAtoi(str);    cout << val;    return 0;}  // This is code is contributed by rathbhupendra
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 <bits/stdc++.h>using namespace std;  // A simple atoi() functionint 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 codeint main(){    char str[] = "-123";      // Function call    int val = myAtoi(str);    cout << val;    return 0;}  // This is code is contributed by rathbhupendra
-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. 
 -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 <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 Codeint main(){    char str[] = "  -123";        // Functional Code    int val = myAtoi(str);    cout <<" "<< val;    return 0;}  // This code is contributed by shivanisinghss2110
 -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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