The parseint Java method is used to simply convert any String to Primitive int type.

An int basically means Integer (…-3,-2,-1,0,1,2,3…) with a range of -2147483648 to 2147483647.

Syntax:

1. `Integer.parseInt(String mydata);`

2. `Integer.parseInt(String mydata, int radix);`

In a shell, there are two Java syntaxes to use `parseInt()` method. It takes in the provided String data as the first parameter. If a second parameter is provided it should be of int type denoting the radix.

A radix is the base of a number. For instance, 2 means Binary, 8 means Octal, 10 is Decimal and 16 is hexadecimal. Imagine radix as the number of unique digits one can use including 0 to represent numbers.

### Example 1

Convert a String to Integer and return the square of the number.

Solution

``````public class ParseInt {
public static void main(String[] args) {
String s = "11";
int i = Integer.parseInt(s); // conversion
int result = i*i;
System.out.println("The square of the number is: "+result);
}
}``````

Output
`The square of the number is: 121`

### Java parseint Example 2

In a school, a student with id 1 has left. Now modify all the student ids(reduce by 1). You are given a String of a student name with the last 2 characters as id number. Extract the id number. Display the name with the correct id.

Sample
`input: John65 Output: name-John roll-64`

`input: Ana03 Output: name-Ana roll-2`

Solution

``````public class ParseInt {
public static void main(String[] args) {
String s = "John65";
String name = s.substring(0, s.length()-2);//extract name
String id_string = s.substring(s.length()-2);//extract id
int id_int = Integer.parseInt(id_string);//convert id from String to int
int correct_id = id_int-1;//correct id
System.out.println("name-"+name+" roll-"+correct_id);//print output
}
}
``````

Output
`name-John roll-64`

### Exceptions in parseint Java

NumberFormatException is generally thrown by the `parseInt()` method in Java, if any of the following cases occur:

1. The string is null or empty
2. In addition, The value in the string is not an integer
3. Specifically for the `parseInt(String s, int radix)` a variant of the function, if the number contains digits outside the range of the specified radix. For instance, if we define radix as 2 and pass “101” and “121”. Clearly, radix 2 means it is a Binary number. We know that 121 is not a valid Binary number. Let’s code and see.

Valid Case – Example Program

``````public class ParseInt {
public static void main(String[] args) {
String s = "101";
System.out.println(Integer.parseInt(s,2));
}
}
``````

Output Valid Output of parseInt() Java

Invalid Case – Example Program

``````public class ParseInt {
public static void main(String[] args) {
String s = "121";
System.out.println(Integer.parseInt(s,2));
}
}
``````

Output Invalid Output of parseInt() Java

Let’s say we want to handle the above exception. Instead of getting the entire program to a halt, we want to tell the users about the problem.

The exercise is simple. If a NumberFormatException occurs, tell the user that it is not a valid number!

Great if you tried! Here’s the solution for help.

code

``````public class ParseInt {
public static void main(String[] args) {
try {
String s = "%101"; // invalid number
System.out.println(Integer.parseInt(s));
}
catch(NumberFormatException e) {
System.out.println("not a valid number!");
}
}
}
``````

Output
`not a valid number!`

Finally, We have used a try-catch block. If a NumberFormatException occurs the catch block is executed. Then we can tell users about the exception.

Likewise, You can also read out popular posts on int to String Java – 5 Ways to Convert int to String in Java

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