字节组合生成器和定序器

Question

首先，这个问题将把关于同一代码的 2 个问题合并为一个。另外，请原谅我对语言的无知，我最近两周才学 Java，这是我第一个主要项目的一部分。

考虑以下代码：

public class TESTCODE {

public static ArrayList<String> bytePossibalitiyGenerator(int bits, String current) throws Exception {
    ArrayList<String> binaries = new ArrayList<>();
    if (bits%8 != 0) {
        int crash = bits%8;
        throw new Exception("The bit count that you have entered is not divisable by 8:" + "\n" + "There is a remainder of: " + Integer.toString(crash));
    } else {
        if (current.length() == bits) {
            binaries.add(current);
            return binaries;
        }
        // pad a 0 and 1 in front of current;
        binaries.addAll(bytePossibalitiyGenerator(bits, "0" + current));
        binaries.addAll(bytePossibalitiyGenerator(bits, "1" + current));
        System.out.println(binaries.toString());
    }
    return binaries;
}

//The method below is supposed to format out the whitespace between binary strings and arrange the 
//data in such a way that each possible outcome is on a new line.
    //TODO  Add a parser for the size of the byte ex. if the binary string is comprised of all of the possible
    //TODO  outcomes for 1 byte than every 8 instead of appending a space, append a new line.

public static String binarySequencer(String input) {
    StringBuffer toReturn = new StringBuffer();
    //This StringBuilder is a safety precaution to ensure that if the algorithm is to be 
    //run again, the value of each previously read and appended string position is nullified
    //so that it is not re-appended to the StringBuffer
    StringBuilder inputSB = new StringBuilder(input);

    //Setting the booleans for which type of string the input was (The raw binary array itself, or the array converted into a string)
    boolean rawBinaryArrayOutput = Pattern.compile("^[0-1,\s]+$").matcher(input).find();
    boolean stringBinary = Pattern.compile("^[0-1\t]+$").matcher(input).find();

    //This boolean is to check whether the loop has previously passed over 1 byte for sequencing
    boolean hasPassedAByte = false;

    //Safety if statements, because who doesn't love Java Exceptions and stack traces...
    if (rawBinaryArrayOutput == false && stringBinary == false || rawBinaryArrayOutput == true && stringBinary == true) {
        System.out.println(rawBinaryArrayOutput);
        System.out.println(stringBinary);
        //TODO Find a way to print the stack trace...
        throw new InputMismatchException();
    } else {
        int runLength = 0;
        for (int i = 0; i < inputSB.length(); i++) {
            int j = 0;
            if (stringBinary == true && rawBinaryArrayOutput == false) {
                if (runLength == 8 && (inputSB.charAt(j += i) == 0 || inputSB.charAt(j += 1) == 1)) {
                    toReturn.append(Character.toString(' '));
                    runLength = 0;
                    hasPassedAByte = true;
                } else {
                    if (hasPassedAByte = true && runLength == 8 && (inputSB.charAt(j) != 0 || inputSB.charAt(j) != 1)) {
                        toReturn.append("\n");
                        runLength = 0;
                        hasPassedAByte = false;
                    }
                    while (i + 1 < inputSB.length() && (inputSB.charAt(i) == 0 || inputSB.charAt(i) == 1) && runLength != 8) {
                        runLength++;
                        toReturn.append(inputSB.charAt(i));
                        inputSB.insert(i, null);
                        i++;
                    }

                }
            } else {
                if (rawBinaryArrayOutput == true && stringBinary == false) {
                    //Insert code for formatting the raw binary array output
                    System.out.println("You haven't added this code yet :p");

                }
            }
        }
    }
    return toReturn.toString();
}

public static void main(String[] args) throws Exception {
String toBeSequenced = "";
for (String s : bytePossibalitiyGenerator(16, "")) {
    toBeSequenced += s + "\t";
}
System.out.println(binarySequencer(toBeSequenced));}}

现在开始提问：

1：对于 boolean rawBinaryArrayOutput 我正在使用 java.util.regex.Pattern class' compile 方法在输入的字符串中搜索 {0-1 , \s} 个字符，并且如果找到其中任何一个，它会将 rawBinaryArrayOutput 设置为 true。如果找到所有这些值中的至少 1 个，有没有办法让它仅将 rawBinaryArrayOutput 设置为 true？

2：在binarySequencer方法中我有一个StringBuilder inputSB自动获取String input的值，这样我就可以修改StringBuilder中的值。在第 75 行，我试图将位置 i 的值设置为 null，这样如果 while 循环以某种方式在同一位置运行两次，它就不会将任何内容附加到 StringBuilder toReturn ，但是 Eclipse 在该行 The method (int, Object) is ambiguous for type StringBuilder 上给我一个编译错误。这是什么意思，我该如何解决？

3：我在 binarySequencer 方法的开头有一个 if 语句，它检查是否 rawBinaryArrayOutput == false && stringBinary == false || rawBinaryArrayOutput == true && stringBinary == true，如果是，那么它将 throw new InputMismatchException();。如果可能的话，我怎样才能让它也打印出堆栈跟踪？

Answer 1

对于 rawBinaryArrayOutput 你可以尝试使用正则表达式：
```
^([01](,\s)?)+$
```
将接受带有 0 或 1 序列的字符串（一个或多个字符）由可选的 ,\s 部分分隔。但是对于位输入，也许更好的是：
```
^(?:(?:[01]{8})+(?:,\s)?)+$
```
将只接受 0 或 8 个字符序列中的一个或多个 1。 ?: 部分用于非正则表达式分组。
The method (int, Object) is ambiguous for type StringBuilder 据我所知，StringBuilder class 有两个相似的方法，在本例中是： insert(int, String) 和 insert(int, char[])，编译器不会知道你想调用哪个，因为作为第二个变量你使用 null，空引用可以转换为任何 class 类型的表达式。尝试：
```
inputSB.insert(i, (char[]) null); 
```
例如。那么很明显你调用了哪个方法，即使它是与案例无关。
我不会用InputMismatchException()，不如再抛一个自定义异常。使用正确的正则表达式，将没有双真匹配选项，使用双假匹配，您可以只打印有用的信息。

字节组合生成器和定序器

Byte combination generator and sequencer

java

regex

binary

recursion

byte