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In this post, we will see about Java.lang.NumberFormatException.
java.lang.NumberFormatException occurs where you try to convert any non-numeric String to Number such as Integer, Double, Float or Long.
This exception generally occurs when you get input from user in the form of String. This is not Java developer‘s fault and mostly data error. In this scenario, data should be corrected and you can log the error to log file for analysis later. If you do not handle the application then the application might crash.
Let’s understand it with the help of an example.
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package org.arpit.java2blog; import java.util.Scanner; public class StringToIntegerConversion { public static void main(String[] args) { Scanner scanner = null ; try { System.out.println("Please enter a number:"); scanner = new Scanner(System.in); String inputStr = scanner.nextLine(); int i = Integer.parseInt(inputStr); //If NumberFormatException occurs, this will become unreachable code System.out.println("Converted int value = " + i); } catch (NumberFormatException nfe) { nfe.printStackTrace(); } finally { scanner.close(); } } } |
When you run above program, you will get below output:
Iteration 1: Put correct Integer value
10
Converted int value = 10
Iteration 2: Put NA as user input
NA
java.lang.NumberFormatException: For input string: “NA”
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.base/java.lang.Integer.parseInt(Integer.java:652)
at java.base/java.lang.Integer.parseInt(Integer.java:770)
at org.arpit.java2blog.StringToIntegerConversion.main(StringToIntegerConversion.java:14)
As "NA" can not be converted to Integer, it will throw above exception.
Iteration 3: Put “” as user input
java.lang.NumberFormatException: For input string: “”
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.base/java.lang.Integer.parseInt(Integer.java:662)
at java.base/java.lang.Integer.parseInt(Integer.java:770)
at org.arpit.java2blog.StringToIntegerConversion.main(StringToIntegerConversion.java:14)
As "" can not be converted to Integer, it will throw above exception.
Iteration 4: Put “1.0” as user input
1.0
java.lang.NumberFormatException: For input string: “1.0”
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.base/java.lang.Integer.parseInt(Integer.java:652)
at java.base/java.lang.Integer.parseInt(Integer.java:770)
at org.arpit.java2blog.StringToIntegerConversion.main(StringToIntegerConversion.java:14)
It is also necessay that given input string should be able to convert to target data type as "1.0" can not be converted to Integer.
How to resolve java.lang.NumberFormatException
If you are getting java.lang.NumberFormatException, you can easily find information in exception itself.
java.lang.NumberFormatException: For input string: “NA”
As NA is alphanumeric and can not be converted to Integer.
Only way to avoid this exception is correct the input data and you should put proper exception handling whenever you are doing String to Number conversion.
