字节组合发生器和序列器

首先，这个问题将是关于相同代码的两个问题。另外，请原谅我对这门语言的无知，我在过去的两周才学会Java，这是我第一个主要项目的一部分。

考虑以下代码:

<>之前公共类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));}}

之前

现在是问题:

1:对于boolean rawBinaryArrayOutput，我使用java.util.regex.Pattern类的compile方法来搜索正在输入的字符串中的{0-1 , \s}字符，如果它找到其中的任何一个，它将rawBinaryArrayOutput设置为true。是否有一种方法，使它只设置rawBinaryArrayOutput为真，如果发现所有这些值中的至少一个?

2:在binarySequencer方法中，我有一个StringBuilder inputSB自动取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，以便如果while循环以某种方式在同一位置上运行两次，它不会向StringBuilder toReturn追加任何内容，但是Eclipse在The method (int, Object) is ambiguous for type StringBuilder行上给了我一个编译错误。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，检查是否rawBinaryArrayOutput == false && stringBinary == false || rawBinaryArrayOutput == true && stringBinary == true，如果它们是，那么它将throw new InputMismatchException();。如果可能的话，我该如何让它也打印出堆栈跟踪呢?

对于rawBinaryArrayOutput，您可以尝试使用regex:
```
^([01](,s)?)+$
```
将接受具有0或1序列的字符串(一个或多个char)由可选的,s部分分隔。但是对于比特输入，可能最好是:
```
^(?:(?:[01]{8})+(?:,s)?)+$
```
只接受0或8个字符序列中的一个或多个1。

?:

The method (int, Object) is ambiguous for type StringBuilder据我所知，意味着StringBuilder类有两个类似的方法，在这种情况下，它是:insert(int, String)和insert(int, char[])，以及编译器没有知道要调用哪个变量，因为作为第二个变量使用null, null引用可以转换为任何类类型的表达式。试一试:
```
inputSB.insert(i, (char[]) null); 
```
例如
。那么你调用哪个方法就很明显了，即使它是
我不会使用InputMismatchException()，最好抛出另一个自定义异常。使用正确的正则表达式，将没有双真匹配的选项，而使用双假匹配，您可以只打印有用的信息。

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