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Java Long decode() method with Examples

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The java.lang.Long.decode() is a built in function in java that decodes a String into a Long. It accepts decimal, hexadecimal, and octal numbers.

Syntax:

public static Long decode(String number) throws NumberFormatException 

Parameter: 
number-  the number that has to be decoded into a Long. 

Error and Exceptions:

  • NumberFormatException: If the String does not contain a parsable long, the program returns this error.

Returns: It returns the decoded string.

Program 1: The program below demonstrates the working of function.




// Java program to demonstrate
// of java.lang.Long.decode() method
import java.lang.Math;
  
class GFG {
  
    // driver code
    public static void main(String args[])
    {
  
        // demonstration of function
        Long l = new Long(14);
        String str = "54534";
  
        System.out.println("Number = "
                        + l.decode(str));
    }
}


Output:

Number = 54534 

Program 2: The program demonstrates the conversions using decode() function




// Java program to demonstrate
// of java.lang.Long.decode() method
import java.lang.Math;
  
class GFG {
  
    // driver code
    public static void main(String args[])
    {
        // demonstration of conversions
        String decimal = "10"; // Decimal
        String hexa = "0XFF"; // Hexa
        String octal = "067"; // Octal
  
        // convert decimal val to number using decode() method
        Integer number = Integer.decode(decimal);
        System.out.println("Decimal [" + decimal + "] = " + number);
  
        number = Integer.decode(hexa);
        System.out.println("Hexa [" + hexa + "] = " + number);
  
        number = Integer.decode(octal);
        System.out.println("Octal [" + octal + "] = " + number);
    }
}


Output:

Decimal [10] = 10
Hexa [0XFF] = 255
Octal [067] = 55

Program 3: The program demonstrates error and exceptions.




// Java program to demonstrate
// of java.lang.Long.decode() method
import java.lang.Math;
  
class GFG {
  
    // driver code
    public static void main(String args[])
    {
        // demonstration of errorand exception when
        // a non-parsable Long is passed
  
        String decimal = "1A";
  
        // throws an error
        Integer number = Integer.decode(decimal);
        System.out.println("string [" + decimal + "] = " + number);
    }
}


Output:

 
Exception in thread "main" java.lang.NumberFormatException: For input string: "1A"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:580)
    at java.lang.Integer.valueOf(Integer.java:740)
    at java.lang.Integer.decode(Integer.java:1197)
    at GFG.main(File.java:16)



Last Updated : 05 Dec, 2018
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