我想将一个十六进制字符串转换为二进制字符串。例如,十六进制数2对应的二进制数为0010。以下是代码:
String HexToBinary(String Hex)
{
int i = Integer.parseInt(Hex);
String Bin = Integer.toBinaryString(i);
return Bin;
}
然而,这仅适用于十六进制 0 - 9;对于十六进制 A - F,它不起作用,因为它使用了int
。是否有人可以改进它?
我想将一个十六进制字符串转换为二进制字符串。例如,十六进制数2对应的二进制数为0010。以下是代码:
String HexToBinary(String Hex)
{
int i = Integer.parseInt(Hex);
String Bin = Integer.toBinaryString(i);
return Bin;
}
然而,这仅适用于十六进制 0 - 9;对于十六进制 A - F,它不起作用,因为它使用了int
。是否有人可以改进它?
你需要告诉Java这个整数是十六进制的,像这样:
String hexToBinary(String hex) {
int i = Integer.parseInt(hex, 16);
String bin = Integer.toBinaryString(i);
return bin;
}
接受的版本仅适用于32位数字。
这里有一个可以处理任意长度十六进制字符串的版本:
public static String hexToBinary(String hex) {
return new BigInteger(hex, 16).toString(2);
}
被接受的答案仅适用于32位值,而替代的BigInteger版本会截断二进制字符串中的前导零!这里有一个在所有情况下应该都有效的函数。
public static String hexToBinary(String hex) {
int len = hex.length() * 4;
String bin = new BigInteger(hex, 16).toString(2);
//left pad the string result with 0s if converting to BigInteger removes them.
if(bin.length() < len){
int diff = len - bin.length();
String pad = "";
for(int i = 0; i < diff; ++i){
pad = pad.concat("0");
}
bin = pad.concat(bin);
}
return bin;
}
这是我为操作十六进制、纯文本和二进制编写的一些例程,希望它们有所帮助。由于我从这些主题中借鉴了一些想法,所以想分享一下。
public static String zero_pad_bin_char(String bin_char){
int len = bin_char.length();
if(len == 8) return bin_char;
String zero_pad = "0";
for(int i=1;i<8-len;i++) zero_pad = zero_pad + "0";
return zero_pad + bin_char;
}
public static String plaintext_to_binary(String pt){
return hex_to_binary(plaintext_to_hex(pt));
}
public static String binary_to_plaintext(String bin){
return hex_to_plaintext(binary_to_hex(bin));
}
public static String plaintext_to_hex(String pt) {
String hex = "";
for(int i=0;i<pt.length();i++){
String hex_char = Integer.toHexString(pt.charAt(i));
if(i==0) hex = hex_char;
else hex = hex + hex_char;
}
return hex;
}
public static String binary_to_hex(String binary) {
String hex = "";
String hex_char;
int len = binary.length()/8;
for(int i=0;i<len;i++){
String bin_char = binary.substring(8*i,8*i+8);
int conv_int = Integer.parseInt(bin_char,2);
hex_char = Integer.toHexString(conv_int);
if(i==0) hex = hex_char;
else hex = hex+hex_char;
}
return hex;
}
public static String hex_to_binary(String hex) {
String hex_char,bin_char,binary;
binary = "";
int len = hex.length()/2;
for(int i=0;i<len;i++){
hex_char = hex.substring(2*i,2*i+2);
int conv_int = Integer.parseInt(hex_char,16);
bin_char = Integer.toBinaryString(conv_int);
bin_char = zero_pad_bin_char(bin_char);
if(i==0) binary = bin_char;
else binary = binary+bin_char;
//out.printf("%s %s\n", hex_char,bin_char);
}
return binary;
}
public static String hex_to_plaintext(String hex) {
String hex_char;
StringBuilder plaintext = new StringBuilder();
char pt_char;
int len = hex.length()/2;
for(int i=0;i<len;i++){
hex_char = hex.substring(2*i,2*i+2);
pt_char = (char)Integer.parseInt(hex_char,16);
plaintext.append(pt_char);
//out.printf("%s %s\n", hex_char,bin_char);
}
return plaintext.toString();
}
}
private static Map<String, String> digiMap = new HashMap<>();
static {
digiMap.put("0", "0000");
digiMap.put("1", "0001");
digiMap.put("2", "0010");
digiMap.put("3", "0011");
digiMap.put("4", "0100");
digiMap.put("5", "0101");
digiMap.put("6", "0110");
digiMap.put("7", "0111");
digiMap.put("8", "1000");
digiMap.put("9", "1001");
digiMap.put("A", "1010");
digiMap.put("B", "1011");
digiMap.put("C", "1100");
digiMap.put("D", "1101");
digiMap.put("E", "1110");
digiMap.put("F", "1111");
}
static String hexToBin(String s) {
char[] hex = s.toCharArray();
String binaryString = "";
for (char h : hex) {
binaryString = binaryString + digiMap.get(String.valueOf(h));
}
return binaryString;
}
对不起,回复有点晚了。但是我认为我的答案是最直接和简单的。
还请理解。