带有示例的Java Long decode() 方法

Java .lang.Long.decode()是Java中的内置函数，可将String 解码为Long。它接受十进制、十六进制和八进制数。

句法：

public static Long decode(String number) throws NumberFormatException 

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

错误和异常：

• NumberFormatException：如果 String 不包含可解析的 long，则程序返回此错误。

返回：它返回解码后的字符串。

程序1：下面的程序演示了函数的工作。

// 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));
    }
}

输出：

Number = 54534 

程序 2：程序演示了使用 decode()函数的转换

// 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);
    }
}

输出：

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

程序 3：程序演示错误和异常。

// 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);
    }
}

输出：

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)