Java - Integer parseInt() method

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Description

The Java Integer parseInt(String s) method parses the string argument s as a signed decimal integer.

Declaration

Following is the declaration for java.lang.Integer.parseInt() method

 public static int parseInt(String s) throws NumberFormatException 

Parameters

s − This is a String containing the int representation to be parsed.

Return Value

This method returns the integer value represented by the argument in decimal.

Exception

NumberFormatException − if the string does not contain a parsable integer.

Getting an Integer from a String having positive int value Example

The following example shows the usage of Integer parseInt() method to parse a Integer object from a string containing decimal number. We've created a String variable and assign it a string containing decimal number. Then using parseInt method, we're obtaining the Integer object and printing it.

 package com.tutorialspoint; public class IntegerDemo { public static void main(String[] args) { String str = "50"; /* returns an Integer object holding the int value represented by string str */ System.out.println("Number = " + Integer.parseInt(str)); } } 

Output

Let us compile and run the above program, this will produce the following result −

 Number = 50 

Getting an Integer from a String having negative int value Example

The following example shows the usage of Integer parseInt() method to get a Integer object from a string containing a negative decimal number. We've created a String variable and assign it a string containing a negative decimal number. Then using parseInt method, we're obtaining the Integer object and printing it.

 package com.tutorialspoint; public class IntegerDemo { public static void main(String[] args) { String str = "-50"; /* returns an Integer object holding the int value represented by string str */ System.out.println("Number = " + Integer.parseInt(str)); } } 

Output

Let us compile and run the above program, this will produce the following result −

 Number = -50 

Facing Exception while Getting an Integer from a String having octal value Example

The following example shows the usage of Integer parseInt() method to get a Integer object from a string containing a octal number. We've created a String variable and assign it a string containing an octal number. Then using parseInt method, we're trying to obtain an Integer object and exception will be raised.

 package com.tutorialspoint; public class IntegerDemo { public static void main(String[] args) { String str = "0x3"; /* returns an Integer object holding the int value represented by string str */ System.out.println("Number = " + Integer.parseInt(str)); } } 

Output

Let us compile and run the above program, this will produce the following result −

 Exception in thread "main" java.lang.NumberFormatException: For input string: "0x3"	at java.lang.NumberFormatException.forInputString(Unknown Source)	at java.lang.Integer.parseInt(Unknown Source)	at java.lang.Integer.parseInt(Unknown Source)	at com.tutorialspoint.IntegerDemo.main(IntegerDemo.java:10)