C++ Program To Write Your Own atoi()
Last Updated :
18 Aug, 2023
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
OutputString 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!
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