parseInt Java Method | Implementation with code example

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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