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 of atoi()
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[] = "neveropen";
val = atoi(strn2);
cout << "String value = " << strn2 << endl;
cout << "Integer value = " << val << endl;
return (0);
}
|
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[] = "neveropen";
val = atoi(strn2);
printf("String value = %s\n", strn2);
printf("Integer value = %d\n", val);
return (0);
}
|
Java
import java.util.*;
public class Main {
public static void main(String[] args)
{
int val;
String strn1 = "12546";
val = Integer.parseInt(strn1);
System.out.println("String value = " + strn1);
System.out.println("Integer value = " + val);
String strn2 = "neveropen";
try {
val = Integer.parseInt(strn2);
System.out.println("String value = " + strn2);
System.out.println("Integer value = " + val);
}
catch (NumberFormatException e) {
val = 0;
System.out.println("String value = " + strn2);
System.out.println("Integer value = " + val);
}
}
}
|
Python3
def main():
strn1 = "12546"
val = int(strn1)
print("String value = ", strn1)
print("Integer value = ", val)
strn2 = "neveropen"
try:
val = int(strn2)
except ValueError:
val = 0
print("String value = ", strn2)
print("Integer value = ", val)
if __name__ == "__main__":
main()
|
C#
using System;
namespace neveropen {
class Program {
static void Main(string[] args)
{
String strn1 = "12546";
int val = int.Parse(strn1);
Console.WriteLine("String value = " + strn1);
Console.WriteLine("Integer value = " + val);
String strn2 = "neveropen";
try {
val = int.Parse(strn2);
}
catch (FormatException e) {
val = 0;
Console.WriteLine("String value = " + strn2);
Console.WriteLine("Integer value = " + val);
}
}
}
}
|
Javascript
let val;
let strn1 = "12546";
val = parseInt(strn1);
console.log("String value = " + strn1);
console.log("Integer value = " + val);
let strn2 = "neveropen";
val = parseInt(strn2);
console.log("String value = " + strn2);
console.log("Integer value = " + val);
|
Output
String value = 12546
Integer value = 12546
String value = neveropen
Integer value = 0
Complexity Analysis:
- Time Complexity: O(n), Only one traversal of the string is needed.
- Space Complexity: O(1), As no extra space is required.
Now let’s understand various ways in which one can create their own atoi() function supported by various conditions:
Approach 1
The following is a simple implementation of conversion without considering any special case.
- Initialize the result as 0.
- Start from the first character and update the result for every character.
- For every character update the answer as result = result * 10 + (s[i] – ‘0’)
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int myAtoi(char* str)
{
int res = 0;
for (int i = 0; str[i] != '\0'; ++i)
res = res * 10 + str[i] - '0';
return res;
}
int main()
{
char str[] = "89789";
int val = myAtoi(str);
cout << val;
return 0;
}
|
C
#include <stdio.h>
int myAtoi(char* str)
{
int res = 0;
for (int i = 0; str[i] != '\0'; ++i)
res = res * 10 + str[i] - '0';
return res;
}
int main()
{
char str[] = "89789";
int val = myAtoi(str);
printf("%d ", val);
return 0;
}
|
Java
class GFG {
static int myAtoi(String str)
{
int res = 0;
for (int i = 0; i < str.length(); ++i)
res = res * 10 + str.charAt(i) - '0';
return res;
}
public static void main(String[] args)
{
String str = "89789";
int val = myAtoi(str);
System.out.println(val);
}
}
|
Python3
def myAtoi(string):
res = 0
for i in range(len(string)):
res = res * 10 + (ord(string[i]) - ord('0'))
return res
string = "89789"
print(myAtoi(string))
|
C#
using System;
class GFG {
static int myAtoi(string str)
{
int res = 0;
for (int i = 0; i < str.Length; ++i)
res = res * 10 + str[i] - '0';
return res;
}
public static void Main()
{
string str = "89789";
int val = myAtoi(str);
Console.Write(val);
}
}
|
Javascript
<script>
function myAtoi(str)
{
let res = 0;
for (let i = 0; i < str.length; ++i)
res = res * 10 + str[i].charCodeAt(0) - '0'.charCodeAt(0);
return res;
}
let str = "89789";
let val = myAtoi(str);
document.write(val);
</script>
|
Complexity Analysis:
- Time Complexity: O(n), Only one traversal of the string is needed.
- Space Complexity: O(1), As no extra space is required.
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 the sign with it.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int myAtoi(char* str)
{
int res = 0;
int sign = 1;
int i = 0;
if (str[0] == '-') {
sign = -1;
i++;
}
for (; str[i] != '\0'; i++)
res = res * 10 + str[i] - '0';
return sign * res;
}
int main()
{
char str[] = "-123";
int val = myAtoi(str);
cout << val;
return 0;
}
|
C
#include <stdio.h>
int myAtoi(char* str)
{
int res = 0;
int sign = 1;
int i = 0;
if (str[0] == '-') {
sign = -1;
i++;
}
for (; str[i] != '\0'; ++i)
res = res * 10 + str[i] - '0';
return sign * res;
}
int main()
{
char str[] = "-123";
int val = myAtoi(str);
printf("%d ", val);
return 0;
}
|
Java
class GFG {
static int myAtoi(char[] str)
{
int res = 0;
int sign = 1;
int i = 0;
if (str[0] == '-') {
sign = -1;
i++;
}
for (; i < str.length; ++i)
res = res * 10 + str[i] - '0';
return sign * res;
}
public static void main(String[] args)
{
char[] str = "-123".toCharArray();
int val = myAtoi(str);
System.out.println(val);
}
}
|
Python3
def myAtoi(string):
res = 0
sign = 1
i = 0
if string[0] == '-':
sign = -1
i += 1
for j in range(i, len(string)):
res = res*10+(ord(string[j])-ord('0'))
return sign * res
string = "-123"
print(myAtoi(string))
|
C#
using System;
class GFG {
static int myAtoi(string str)
{
int res = 0;
int sign = 1;
int i = 0;
if (str[0] == '-') {
sign = -1;
i++;
}
for (; i < str.Length; ++i)
res = res * 10 + str[i] - '0';
return sign * res;
}
public static void Main()
{
string str = "-123";
int val = myAtoi(str);
Console.Write(val);
}
}
|
Javascript
<script>
function myAtoi(str)
{
var res = 0;
var sign = 1;
var i = 0;
if (str[0] == '-') {
sign = -1;
i++;
}
for (; i < str.length; ++i)
res = res * 10 + str[i].charCodeAt(0) - '0'.charCodeAt(0);
return sign * res;
}
var str = "-129";
var val=myAtoi(str);
document.write(val);
</script>
<! --This code is contributed by nirajgusain5 -->
|
Complexity Analysis:
- Time Complexity: O(n), Only one traversal of the string is needed.
- Space Complexity: O(1), As no extra space is required.
Approach 3
Four corner cases need to be handled:
- Discard all leading whitespaces
- Sign of the number
- Overflow
- Invalid Input
Below are the steps for the above approach:
- To remove the leading whitespaces, run a loop and ignore the whitespaces until a character of the digit is reached.
- It keeps a sign variable to keep track of the sign of the number.
- It checks for valid input characters if all characters are from 0 to 9 and converts them into integers.
- If an overflow occurs and if the number is greater than or equal to INT_MAX/10, 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++
#include <bits/stdc++.h>
using namespace std;
int myAtoi(const char* str)
{
int sign = 1, base = 0, i = 0;
while (str[i] == ' ') {
i++;
}
if (str[i] == '-' || str[i] == '+') {
sign = 1 - 2 * (str[i++] == '-');
}
while (str[i] >= '0' && str[i] <= '9') {
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);
cout << " " << val;
return 0;
}
|
C
#include <limits.h>
#include <stdio.h>
int myAtoi(const char* str)
{
int sign = 1, base = 0, i = 0;
while (str[i] == ' ') {
i++;
}
if (str[i] == '-' || str[i] == '+') {
sign = 1 - 2 * (str[i++] == '-');
}
while (str[i] >= '0' && str[i] <= '9') {
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;
}
|
Java
class GFG {
static int myAtoi(char[] str)
{
int sign = 1, base = 0, i = 0;
while (str[i] == ' ') {
i++;
}
if (str[i] == '-' || str[i] == '+') {
sign = 1 - 2 * (str[i++] == '-' ? 1 : 0);
}
while (i < str.length && str[i] >= '0'
&& str[i] <= '9') {
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;
}
public static void main(String[] args)
{
char str[] = " -123".toCharArray();
int val = myAtoi(str);
System.out.printf("%d ", val);
}
}
|
Python3
import sys
def myAtoi(Str):
sign, base, i = 1, 0, 0
while (Str[i] == ' '):
i += 1
if (Str[i] == '-' or Str[i] == '+'):
sign = 1 - 2 * (Str[i] == '-')
i += 1
while (i < len(Str) and
Str[i] >= '0' and Str[i] <= '9'):
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
Str = list(" -123")
val = myAtoi(Str)
print(val)
|
C#
using System;
class GFG {
static int myAtoi(char[] str)
{
int sign = 1, Base = 0, i = 0;
while (str[i] == ' ') {
i++;
}
if (str[i] == '-' || str[i] == '+') {
sign = 1 - 2 * (str[i++] == '-' ? 1 : 0);
}
while (i < str.Length && str[i] >= '0'
&& str[i] <= '9') {
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;
}
public static void Main(String[] args)
{
char[] str = " -123".ToCharArray();
int val = myAtoi(str);
Console.Write("{0} ", val);
}
}
|
Javascript
<script>
function myAtoi(str)
{
var sign = 1, base = 0, i = 0;
while (str[i] == ' ')
{
i++;
}
if (str[i] == '-' || str[i] == '+')
{
sign = 1 - 2 * (str[i++] == '-');
}
while (str[i] >= '0' && str[i] <= '9')
{
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;
}
var str = " -123";
var val = myAtoi(str);
document.write(" ", val);
</script>
|
Complexity Analysis:
- Time Complexity: O(n), Only one traversal of the string is needed.
- Space Complexity: O(1), As no extra space is required.
Related Articles:
Exercise:
Write your won atof() that takes a string (which represents a floating point value) as an argument and returns its value as double.
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