parseInt() Method in Java
Table of Content:
Description
This method is used to get the primitive data type of a certain String. parseXxx() is a static method and can have one argument or two.
Syntax
Following are all the variants of this method ?
static int parseInt(String s) static int parseInt(String s, int radix)
Parameters
Here is the detail of parameters ?
-
s ? This is a string representation of decimal.
-
radix ? This would be used to convert String s into integer.
Return Value
-
parseInt(String s) ? This returns an integer (decimal only).
-
parseInt(int i) ? This returns an integer, given a string representation of decimal, binary, octal, or hexadecimal (radix equals 10, 2, 8, or 16 respectively) numbers as input.
Throws
NumberFormatException : if the string does not contain a parsable integer.
Example
public class ParseIntMethod { public static void main(String args[]) { int x =Integer.parseInt("18"); double c = Double.parseDouble("12"); int b = Integer.parseInt("444",16); System.out.println(x); System.out.println(c); System.out.println(b); } }
This will produce the following result ?
Output
18 12.0 1092 Press any key to continue . . .
Example
public class ParseIntMethod { public static void main(String[] args) { // parsing different strings int z = Integer.parseInt("654",8); int a = Integer.parseInt("-FF", 16); long l = Long.parseLong("2158611234",10); System.out.println(z); System.out.println(a); System.out.println(l); // run-time NumberFormatException will occur here // "Hello" is not a parsable string int x = Integer.parseInt("Hello",8); // run-time NumberFormatException will occur here // (for octal(8),allowed digits are [0-7]) int y = Integer.parseInt("99",8); } }
This will produce the following result ?
Output
428 -255 2158611234 Exception in thread "main" java.lang.NumberFormatException: For input string: "Hello" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:580) at ParseIntMethod.main(ParseIntMethod.java:15) Press any key to continue . . .
Example
public class ParseIntMethod { public static void main(String[] args) { // parsing different strings int z = Integer.parseInt("654"); long l = Long.parseLong("2158611234"); System.out.println(z); System.out.println(l); // run-time NumberFormatException will occur here // "Hello" is not a parsable string int x = Integer.parseInt("Hello"); // run-time NumberFormatException will occur here // (for decimal(10),allowed digits are [0-9]) int a = Integer.parseInt("-FF"); } }
This will produce the following result ?
Output
654 2158611234 Exception in thread "main" java.lang.NumberFormatException: For input string: "Hello" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:580) at java.lang.Integer.parseInt(Integer.java:615) at ParseIntMethod.main(ParseIntMethod.java:13) Press any key to continue . . .