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:
#include <stdio.h> #include <stdlib.h> #include <string.h> int main() { int val; char strn1[] = "12546"; val = atoi(strn1); printf("String value = %s\n", strn1); printf("Integer value = %d\n", val); char strn2[] = "GeeksforGeeks"; val = atoi(strn2); printf("String value = %s\n", strn2); printf("Integer value = %d\n", val); return (0); } |
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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 for (int i = 0; str[i] != '\0'; ++i) res = res * 10 + str[i] - '0'; // return result. return res; } // Driver program to test above function int main() { char str[] = "89789"; int val = myAtoi(str); printf("%d ", val); return 0; } |
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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 for (int i = 0; str[i] != '\0'; ++i) res = res * 10 + str[i] - '0'; // return result. return res; } // Driver code int main() { char str[] = "89789"; int val = myAtoi(str); cout << val; return 0; } // This is code is contributed by rathbhupendra |
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Java
// A simple Java program for // implementation of atoi class GFG { // A simple atoi() function static int myAtoi(String str) { // Initialize result int res = 0; // Iterate through all characters of // input string and update result for (int i = 0; i < str.length(); ++i) res = res * 10 + str.charAt(i) - '0'; // return result. return res; } // Driver code public static void main(String[] args) { String str = "89789"; int val = myAtoi(str); System.out.println(val); } } // This code is contributed by PrinciRaj1992 |
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Python
# Python program for implementation of atoi # A simple atoi() function def myAtoi(string): res = 0 # Iterate through all characters of input string and # update result for i in xrange(len(string)): res = res * 10 + (ord(string[i]) - ord('0')) return res # Driver program string = "89789"print myAtoi(string) # This code is contributed by BHAVYA JAIN |
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C#
// A simple C# program for implementation // of atoi using System; class GFG { // A simple atoi() function static int myAtoi(string str) { int res = 0; // Initialize result // Iterate through all characters of // input string and update result for (int i = 0; i < str.Length; ++i) res = res * 10 + str[i] - '0'; // return result. return res; } // Driver program to test above function public static void Main() { string str = "89789"; int val = myAtoi(str); Console.Write(val); } } // This code is contributed by Sam007. |
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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 <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] != '\0'; ++i) res = res * 10 + str[i] - '0'; // Return result with sign return sign * res; } // Driver code int main() { char str[] = "-123"; int val = myAtoi(str); cout << val; return 0; } // This is code is contributed by rathbhupendra |
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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] != '\0'; ++i) res = res * 10 + str[i] - '0'; // Return result with sign return sign * res; } // Driver program to test above function int main() { char str[] = "-123"; int val = myAtoi(str); printf("%d ", val); return 0; } |
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Java
// Java program for // implementation of atoi class GFG { // A simple atoi() function static 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 (; i < str.length; ++i) res = res * 10 + str[i] - '0'; // Return result with sign return sign * res; } // Driver code public static void main(String[] args) { char[] str = "-123".toCharArray(); int val = myAtoi(str); System.out.println(val); } } // This code is contributed by 29AjayKumar |
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Python
# Python program for implementation of atoi # A simple atoi() function def myAtoi(string): res = 0 # initialize sign as positive sign = 1 i = 0 # if number is negative then update sign if string[0] == '-': sign = -1 i+= 1 # Iterate through all characters of input string and update result for j in xrange(i, len(string)): res = res * 10 + (ord(string[j]) - ord('0')) return sign * res # Driver program string = "-123"print myAtoi(string) # This code is contributed by BHAVYA JAIN |
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C#
// C# program for implementation of atoi using System; class GFG { // A simple atoi() function static int myAtoi(string 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 (; i < str.Length; ++i) res = res * 10 + str[i] - '0'; // Return result with sign return sign * res; } // Driver program to test above function public static void Main() { string str = "-123"; int val = myAtoi(str); Console.Write(val); } } // This code is contributed by Sam007. |
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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.
C++
// A simple C++ program for // implementation of atoi #include <stdio.h> // A utility function to check // whether x is numeric bool isNumericChar(char x) { return (x >= '0' && x <= '9') ? true : false; } // A simple atoi() function. // If the given string contains // any invalid character, then // this function returns 0 int myAtoi(char* str) { if (*str == '\0') return 0; // 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 of // input string and update result for (; str[i] != '\0'; ++i) { // You may add some lines // to write error message if (isNumericChar(str[i]) == false) return 0; // to error stream res = res * 10 + str[i] - '0'; } // Return result with sign return sign * res; } // Driver program to test above function int main() { char str[] = "-134"; int val = myAtoi(str); printf("%d ", val); return 0; } |
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Java
// A simple Java program for // implementation of atoi class Atoi { // A simple atoi() function. // If the given string contains // any invalid character, then // this function returns 0 public static int atoi(String str) { if (str.length() == 0) { return -1; } // Initialize sign as positive int sign = 1; // Initialize result int res = 0; // Initialize index of first digit int i = 0; // If number is negative, // then update sign if (str.charAt(i) == '-') { sign = -1; // Also update index of first digit i++; } // Iterate through all digits of input // string and update result for (; i < str.length(); i++) { if (Character.isDigit(str.charAt(i)) == false) { return -1; } res = res * 10 + Character.getNumericValue( str.charAt(i)); } return sign * res; } // Driver code public static void main(String[] args) { String str = "-134"; int num = atoi(str); System.out.println(num); } } // This code is contributed by Anonymous_Geek |
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Python
# Python program for implementation of atoi # A utility function to check whether x is numeric def isNumericChar(x): if (x >= '0' and x <= '9'): return True return False # A simple atoi() function. If the given string contains # any invalid character, then this function returns 0 def myAtoi(string): if len(string) == 0: return 0 res = 0 # initialize sign as positive sign = 1 i = 0 # if number is negative then update sign if string[0] == '-': sign = -1 i+= 1 # Iterate through all characters of input string and update result for j in xrange(i, len(string)): # You may add some lines to write error message to error stream if isNumericChar(string[j] == False): return 0 res = res * 10 + (ord(string[j]) - ord('0')) return sign * res # Driver program string = "-134"print myAtoi(string) # This code is contributed by BHAVYA JAIN |
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C#
// A simple C# program for implementation of atoi using System; class Atoi { // A simple atoi() function. If the given string contains // any invalid character, then this function returns 0 public static int atoi(String str) { if (str.Length == 0) { return -1; } int sign = 1; // Initialize sign as positive int res = 0; // Initialize result int i = 0; // Initialize index of first digit // If number is negative, then update sign if (str[i] == '-') { sign = -1; i++; // Also update index of first digit } // Iterate through all digits of input // string and update result for (; i < str.Length; i++) { if (char.IsDigit(str[i]) == false) { return -1; } res = res * 10 + int.Parse(str[i].ToString()); } return sign * res; } // Driver code public static void Main(String[] args) { String str = "-134"; int num = atoi(str); Console.WriteLine(num); } } // This code is contributed by 29AjayKumar |
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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; } int main() { char str[] = " -123"; int val = myAtoi(str); printf("%d ", val); return 0; } // This code is contributed by Yogesh shukla. |
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Java
// A simple Java program for // implementation of atoi class GFG { static int myAtoi(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++] == '-' ? 1 : 0); } // checking for valid input while (i < str.length && str[i] >= '0' && str[i] <= '9') { // handling overflow test case if (base > Integer.MAX_VALUE / 10 || (base == Integer.MAX_VALUE / 10 && str[i] - '0' > 7)) { if (sign == 1) return Integer.MAX_VALUE; else return Integer.MIN_VALUE; } base = 10 * base + (str[i++] - '0'); } return base * sign; } // Driver code public static void main(String[] args) { char str[] = " -123".toCharArray(); int val = myAtoi(str); System.out.printf("%d ", val); } } // This code is contributed by 29AjayKumar |
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C#
// A simple C# program for implementation of atoi using System; class GFG { static int myAtoi(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++] == '-' ? 1 : 0); } // checking for valid input while ( i < str.Length && str[i] >= '0' && str[i] <= '9') { // handling overflow test case if (Base > int.MaxValue / 10 || (Base == int.MaxValue / 10 && str[i] - '0' > 7)) { if (sign == 1) return int.MaxValue; else return int.MinValue; } Base = 10 * Base + (str[i++] - '0'); } return Base * sign; } // Driver code public static void Main(String[] args) { char[] str = " -123".ToCharArray(); int val = myAtoi(str); Console.Write("{0} ", val); } } // This code is contributed by 29AjayKumar |
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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.
Exercise:
Write your won atof() that takes a string (which represents an floating point value) as an argument and returns its value as double.
This article is compiled by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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