Java 长解码()方法示例

原文:https://www . geesforgeks . org/Java-long-decode-method-with-examples/

java.lang.Long.decode() 是 java 中的一个内置函数，它将一个字符串解码成一个 Long。它接受十进制、十六进制和八进制数字。

语法:

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: 如果字符串不包含可解析的长字符串，程序将返回此错误。

返回:返回解码后的字符串。

程序 1: 下面的程序演示了功能的工作。

// 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: 程序使用 decode()函数演示转换

// 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)

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