我如何将long转换为其32位二进制表示形式？

Question

我正在执行以下编程练习：int32 to IPv4。该语句是：

采用以下IPv4地址：128.32.10.1🌐

此地址有4个八位位组，其中每个八位位组是一个字节（或8个位）。

1st octet 128 has the binary representation: 10000000
2nd octet 32 has the binary representation: 00100000
3rd octet 10 has the binary representation: 00001010
4th octet 1 has the binary representation: 00000001

所以128.32.10.1 == 10000000.00100000.00001010.00000001

由于上述IP地址具有32位，因此我们可以将其表示为无符号的32位数字：2149583361

完成使用无符号32位数字并返回的函数其IPv4地址的字符串表示形式。范例

2149583361 ==>“ 128.32.10.1” 32 ==>“ 0.0.0.32” 0 ==>“ 0.0.0.0”

首先，我尝试了以下代码：

public class Kata {
    public static String longToIP(long ip) {
        System.out.println("ip: "+ip);
    String binary = Long.toBinaryString(ip);
    System.out.println("binary: "+binary);
    return String.format("%s.%s.%s.%s",Long.parseLong(binary.substring(0,8),2),Long.parseLong(binary.substring(8,16),2),
            Long.parseLong(binary.substring(16,24),2),Long.parseLong(binary.substring(24),2));
    }
}

并且正在测试：

import org.junit.Test;

import java.util.Random;

import static org.junit.Assert.assertEquals;

public class KataTest {
    @Test
    public void sampleTest() {
        assertEquals("128.114.17.104", Kata.longToIP(2154959208L));
        assertEquals("0.0.0.0", Kata.longToIP(0));
        assertEquals("128.32.10.1", Kata.longToIP(2149583361L));
    }
}

[当输入为零时，我的代码引发异常：

java.lang.StringIndexOutOfBoundsException: begin 0, end 8, length 1
at Kata.longToIP(Kata.java:6)

由于在第4行，我从long转换为字符串，二进制没有填充零：

ip: 0
binary: 0

我如何将long ip值转换为始终为32位的二进制字符串（添加填充0）？

我尝试了以下方法：

public class Kata {
    public static String longToIP(long ip) {
        System.out.println("ip: "+ip);
    String binary = String.format("%032d",Long.parseLong(Long.toBinaryString(ip)));
    System.out.println("binary: "+binary);
    return String.format("%s.%s.%s.%s",Long.parseLong(binary.substring(0,8),2),Long.parseLong(binary.substring(8,16),2),
            Long.parseLong(binary.substring(16,24),2),Long.parseLong(binary.substring(24),2));
    }
}

并抛出java.lang.NumberFormatException: For input string: "10000000011100100001000101101000" at Kata.longToIP(Kata.java:4)

然后我将其更改为：

public class Kata {
    public static String longToIP(long ip) {
        System.out.println("ip: "+ip);
    String binary = String.format("%032d",Long.parseLong(Long.toBinaryString(ip),2));
    System.out.println("binary: "+binary);
    return String.format("%s.%s.%s.%s",Long.parseLong(binary.substring(0,8),2),Long.parseLong(binary.substring(8,16),2),
            Long.parseLong(binary.substring(16,24),2),Long.parseLong(binary.substring(24),2));
    }
}

而且这也会引发异常：

java.lang.NumberFormatException: For input string: "00000021"
at Kata.longToIP(Kata.java:7)

我读过：

• Converting long to binary in Java gives 31 bits?
• Java - parse binary to long

Answer 1

我们可以添加填充零：

public class Kata {
    public static String longToIP(long ip) {
        System.out.println("ip: "+ip);
    String binary = Long.toBinaryString(ip);
    if(binary.length()<32) binary = "0".repeat(32-binary.length()) + binary;
    System.out.println("binary: "+binary);
    return String.format("%s.%s.%s.%s",Long.parseLong(binary.substring(0,8),2),Long.parseLong(binary.substring(8,16),2),
            Long.parseLong(binary.substring(16,24),2),Long.parseLong(binary.substring(24),2));
    }
}

或者，我们可以在同一句子中使用格式将长字符串转换为二进制字符串：

public class Kata {
    public static String longToIP(long ip) {
        System.out.println("ip: "+ip);
    String binary = String.format("%32s",Long.toBinaryString(ip)).replace(' ','0'); 
    System.out.println("binary: "+binary);
    return String.format("%s.%s.%s.%s",Long.parseLong(binary.substring(0,8),2),Long.parseLong(binary.substring(8,16),2),
            Long.parseLong(binary.substring(16,24),2),Long.parseLong(binary.substring(24),2));
    }
}