Write your own atoi()
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 |
C
#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); } |
String value = 12546 Integer value = 12546 String value = GeeksforGeeks Integer value = 0
Time Complexity: O(1)
Auxiliary Space: O(1)
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() 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] != '\0' ; ++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 |
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 // 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] != '\0' ; ++i) res = res * 10 + str[i] - '0' ; // return result. return res; } // Driver Code int main() { char str[] = "89789" ; // Function call int val = myAtoi(str); printf ( "%d " , val); return 0; } |
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 // 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 ; 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" ; // Function call int val = myAtoi(str); System.out.println(val); } } // This code is contributed by PrinciRaj1992 |
Python3
# 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 range ( len (string)): res = res * 10 + ( ord (string[i]) - ord ( '0' )) return res # Driver program string = "89789" # Function call print (myAtoi(string)) # This code is contributed by BHAVYA JAIN |
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 // 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; i < str.Length; ++i) res = res * 10 + str[i] - '0' ; // return result. return res; } // Driver code public static void Main() { string str = "89789" ; // Function call int val = myAtoi(str); Console.Write(val); } } // This code is contributed by Sam007. |
Javascript
<script> // A simple Javascript program for // implementation of atoi // A simple atoi() function function myAtoi(str) { // Initialize result let 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 (let i = 0; i < str.length; ++i) res = res * 10 + str[i].charCodeAt(0) - '0' .charCodeAt(0); // return result. return res; } // Driver code let str = "89789" ; // Function call let val = myAtoi(str); document.write(val); // This code is contributed by rag2127 </script> |
89789
Time Complexity : O(n) where n is the length of string str.
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" ; // Function call int val = myAtoi(str); cout << val; return 0; } // This is code is contributed by rathbhupendra |
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 code int main() { char str[] = "-123" ; // Function call int val = myAtoi(str); printf ( "%d " , val); return 0; } |
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(); // Function call int val = myAtoi(str); System.out.println(val); } } // This code is contributed by 29AjayKumar |
Python3
# 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 range (i, len (string)): res = res * 10 + ( ord (string[j]) - ord ( '0' )) return sign * res # Driver code string = "-123" # Function call print (myAtoi(string)) # This code is contributed by BHAVYA JAIN |
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 code public static void Main() { string str = "-123" ; // Function call int val = myAtoi(str); Console.Write(val); } } // This code is contributed by Sam007. |
Javascript
<script> // JavaScript program for implementation of atoi // A simple atoi() function function myAtoi(str) { // Initialize result var res = 0; // Initialize sign as positive var sign = 1; // Initialize index of first digit var 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].charCodeAt(0) - '0' .charCodeAt(0); // Return result with sign return sign * res; } // Driver code var str = "-129" ; var val=myAtoi(str); document.write(val); </script> <! --This code is contributed by nirajgusain5 --> |
-123
Time Complexity : O(n) where n is the length of string str.
Approach 3: 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 |
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 Code int main() { char str[] = " -123" ; // Functional Code int val = myAtoi(str); printf ( "%d " , val); return 0; } // This code is contributed by Yogesh shukla. |
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(); // Function call int val = myAtoi(str); System.out.printf( "%d " , val); } } // This code is contributed by 29AjayKumar |
Python3
# A simple Python3 program for # implementation of atoi import sys def myAtoi( Str ): sign, base, i = 1 , 0 , 0 # If whitespaces then ignore. while ( Str [i] = = ' ' ): i + = 1 # Sign of number if ( Str [i] = = '-' or Str [i] = = '+' ): sign = 1 - 2 * ( Str [i] = = '-' ) i + = 1 # Checking for valid input while (i < len ( Str ) and Str [i] > = '0' and Str [i] < = '9' ): # Handling overflow test case if (base > (sys.maxsize / / 10 ) or (base = = (sys.maxsize / / 10 ) and ( Str [i] - '0' ) > 7 )): if (sign = = 1 ): return sys.maxsize else : return - (sys.maxsize) base = 10 * base + ( ord ( Str [i]) - ord ( '0' )) i + = 1 return base * sign # Driver Code Str = list ( " -123" ) # Functional Code val = myAtoi( Str ) print (val) # This code is contributed by divyeshrabadiya07 |
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 |
Javascript
<script> // A simple JavaScript program for // implementation of atoi function myAtoi(str) { var 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 > Number.MAX_VALUE/ 10 || (base == Number.MAX_VALUE / 10 && str[i] - '0' > 7)) { if (sign == 1) return Number.MAX_VALUE; else return Number.MAX_VALUE; } base = 10 * base + (str[i++] - '0' ); } return base * sign; } // Driver code var str = " -123" ; // Function call var val = myAtoi(str); document.write( " " , val); // This code is contributed by shivanisinghss2110 </script> |
-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.
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