首先，这个问题将是关于相同代码打包成一个的2个问题。另外，请原谅我对语言的无知，我在过去两周才学习Java，这是我第一个重大项目的一部分。
请考虑以下代码：
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));}}
现在提出问题：
1：对于boolean rawBinaryArrayOutput我使用java.util.regex.Pattern类'compile方法搜索正在输入的字符串中的{0-1 , \\s}字符，如果找到任何字符，它将rawBinaryArrayOutput设置为true。如果找到所有这些值中的至少一个，有没有办法让它只将rawBinaryArrayOutput设置为true？
2：在binarySequencer方法中，我StringBuilder inputSB会自动获取String input的值，以便我可以修改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某种方式循环在同一位置上运行两次，它就不会向StringBuilder toReturn添加任何内容，但Eclipse在该行The method (int, Object) is ambiguous for type StringBuilder上给出了编译错误。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，用于检查rawBinaryArrayOutput == false && stringBinary == false || rawBinaryArrayOutput == true && stringBinary == true是否为throw new InputMismatchException();，如果它们是myObj = data。如果可能的话，我怎样才能打印出堆栈跟踪？