将十六进制字符串（hex）转换为二进制字符串

Question

我发现将十六进制转换为二进制的方式如下：

String binAddr = Integer.toBinaryString(Integer.parseInt(hexAddr, 16)); 

尽管此方法适用于较小的十六进制数，但如下所示的十六进制数

A14AA1DBDB818F9759

引发NumberFormatException.

因此，我写了以下似乎可行的方法：

private String hexToBin(String hex){
    String bin = "";
    String binFragment = "";
    int iHex;
    hex = hex.trim();
    hex = hex.replaceFirst("0x", "");

    for(int i = 0; i < hex.length(); i++){
        iHex = Integer.parseInt(""+hex.charAt(i),16);
        binFragment = Integer.toBinaryString(iHex);

        while(binFragment.length() < 4){
            binFragment = "0" + binFragment;
        }
        bin += binFragment;
    }
    return bin;
}

上面的方法基本上接受十六进制字符串中的每个字符，如果需要的话，将其转换为等效的二进制零，然后将其连接到返回值。 这是执行转换的正确方法吗？ 还是我忽略了可能导致我的方法失败的事情？

在此先感谢您的协助。

Answer 1

BigInteger.toString(radix)将做您想要的。 只需输入2的基数即可。

static String hexToBin(String s) {
  return new BigInteger(s, 16).toString(2);
}

Answer 2

Integer.parseInt(hex,16);    
System.out.print(Integer.toBinaryString(hex));

将十六进制（String）解析为以16为底的整数，然后使用toBinaryString（int）方法将其转换为二进制字符串

例

int num = (Integer.parseInt("A2B", 16));
System.out.print(Integer.toBinaryString(num));

将打印

101000101011

由Int处理的Max Hex vakue是FFFFFFF

即如果FFFFFFF0通过ti将给出错误

Answer 3

全零：

static String hexToBin(String s) {
    String preBin = new BigInteger(s, 16).toString(2);
    Integer length = preBin.length();
    if (length < 8) {
        for (int i = 0; i < 8 - length; i++) {
            preBin = "0" + preBin;
        }
    }
    return preBin;
}

Answer 4

public static byte[] hexToBin(String str)
    {
        int len = str.length();
        byte[] out = new byte[len / 2];
        int endIndx;

        for (int i = 0; i < len; i = i + 2)
        {
            endIndx = i + 2;
            if (endIndx > len)
                endIndx = len - 1;
            out[i / 2] = (byte) Integer.parseInt(str.substring(i, endIndx), 16);
        }
        return out;
    }

Answer 5

import java.util.*;
public class HexadeciamlToBinary
{
   public static void main()
   {
       Scanner sc=new Scanner(System.in);
       System.out.println("enter the hexadecimal number");
       String s=sc.nextLine();
       String p="";
       long n=0;
       int c=0;
       for(int i=s.length()-1;i>=0;i--)
       {
          if(s.charAt(i)=='A')
          {
             n=n+(long)(Math.pow(16,c)*10);
             c++;
          }
         else if(s.charAt(i)=='B')
         {
            n=n+(long)(Math.pow(16,c)*11);
            c++;
         }
        else if(s.charAt(i)=='C')
        {
            n=n+(long)(Math.pow(16,c)*12);
            c++;
        }
        else if(s.charAt(i)=='D')
        {
           n=n+(long)(Math.pow(16,c)*13);
           c++;
        }
        else if(s.charAt(i)=='E')
        {
            n=n+(long)(Math.pow(16,c)*14);
            c++;
        }
        else if(s.charAt(i)=='F')
        {
            n=n+(long)(Math.pow(16,c)*15);
            c++;
        }
        else
        {
            n=n+(long)Math.pow(16,c)*(long)s.charAt(i);
            c++;
        }
    }
    String s1="",k="";
    if(n>1)
    {
    while(n>0)
    {
        if(n%2==0)
        {
            k=k+"0";
            n=n/2;
        }
        else
        {
            k=k+"1";
            n=n/2;
        }
    }
    for(int i=0;i<k.length();i++)
    {
        s1=k.charAt(i)+s1;
    }
    System.out.println("The respective binary number is : "+s1);
    }
    else
    {
        System.out.println("The respective binary number is : "+n);
    }
  }
}

Answer 6

快速，适用于大字符串：

    private String hexToBin(String hex){
        hex = hex.replaceAll("0", "0000");
        hex = hex.replaceAll("1", "0001");
        hex = hex.replaceAll("2", "0010");
        hex = hex.replaceAll("3", "0011");
        hex = hex.replaceAll("4", "0100");
        hex = hex.replaceAll("5", "0101");
        hex = hex.replaceAll("6", "0110");
        hex = hex.replaceAll("7", "0111");
        hex = hex.replaceAll("8", "1000");
        hex = hex.replaceAll("9", "1001");
        hex = hex.replaceAll("A", "1010");
        hex = hex.replaceAll("B", "1011");
        hex = hex.replaceAll("C", "1100");
        hex = hex.replaceAll("D", "1101");
        hex = hex.replaceAll("E", "1110");
        hex = hex.replaceAll("F", "1111");
        return hex;
    }

Answer 7

public static byte[] hexToBytes(String string) {
 int length = string.length();
 byte[] data = new byte[length / 2];
 for (int i = 0; i < length; i += 2) {
  data[i / 2] = (byte)((Character.digit(string.charAt(i), 16) << 4) + Character.digit(string.charAt(i + 1), 16));
 }
 return data;
}