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[] = "neveropen"; val = atoi(strn2); printf("String value = %s", strn2); printf("Integer value = %d", val); return (0);} |
String value = 12546 Integer value = 12546 String value = neveropen 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() 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); printf("%d ", val); return 0;} |
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() 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); printf("%d ", val); return 0;} |
-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.
-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 Codeint main(){ char str[] = " -123"; // Functional Code int val = myAtoi(str); printf("%d ", val); return 0;}// This code is contributed by Yogesh shukla. |
-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.
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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