如何将String解析为Binary并将其转换为Java中等效的UTF-8? [英] How to parse String as Binary and convert it to UTF-8 equivalent in Java?

问题描述

我需要将String内容解析为二进制序列，并将其转换为等效于 UTF-8 的String.

I need to parse String content as binary sequence and convert them to its UTF-8 equivalent String.

例如， B ， A 和 R 的 UTF-8 二进制等效项如下:
B = 01000010
A = 01000001
R = 01010010

For example, UTF-8 binary equivalents of B, A and R are as follows:
B = 01000010
A = 01000001
R = 01010010

现在，我需要将字符串"010000100100000101010010" 转换为字符串"BAR"
即对于上述情况，将具有24个字符的输入字符串分成三个相等的部分(每部分8个字符)，并转换为与它等效的 UTF-8 字符串值.

Now, I need to convert a string "010000100100000101010010" to string "BAR"
i.e. For above case input string with 24 characters are divided into three equal parts(8 character in each part) and translated to its UTF-8 equivalent as a String value.

示例代码:

public static void main(String args[]) {
    String B = "01000010";
    String A = "01000001";
    String R = "01010010";
    String BAR = "010000100100000101010010";

    String utfEquiv = toUTF8(BAR);//expecting to get "BAR"
    System.out.println(utfEquiv);
}

private static String toUTF8(String str) {
    // TODO 
    return "";
}

方法 toUTF8(String str){}

推荐答案

您应该将其分为两个问题:

You should separate this into two problems:

• 通过解析二进制值将字符串转换为字节数组
• 使用UTF-8将字节数组转换回字符串

使用new String(bytes, StandardCharsets.UTF_8)，后者非常简单.

对于第一部分，棘手的部分是Byte.parseByte不会自动处理前导1 ...因此，我可能会将每个8位字符串解析为short，然后转换为byte :

For the first part, the tricky part is that Byte.parseByte won't automatically handle a leading 1... so I'd probably parse each 8-bit string into a short and then cast to byte:

public static byte[] binaryToBytes(String input) {
    // TODO: Argument validation (nullity, length)
    byte[] ret = new byte[input.length() / 8];
    for (int i = 0; i < ret.length; i++) {
        String chunk = input.substring(i * 8, i * 8 + 8);
        ret[i] = (byte) Short.parseShort(chunk, 2);
    }
    return ret;
}

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