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Java将十六进制值读入int类型的数组

[英]Java reading hex values into an array of type int

I have a file which contains integer numbers represented in hexadecimal IS there any way to store all of these numbers into an array of integers. 我有一个文件,其中包含以十六进制IS表示的整数,可以通过任何方式将所有这些数字存储到整数数组中。

I know you can say int i = 0x 我知道你可以说int i = 0x

but I cannot do this when reading in the values I got an error? 但是在读取出现错误的值时我不能这样做?

Thanks in advance! 提前致谢!

You probably want to go through Integer.parseInt(yourHexValue, 16) . 您可能想要遍历Integer.parseInt(yourHexValue, 16)

Example: 例:

// Your reader
BufferedReader sr = new BufferedReader(new StringReader("cafe\nBABE"));

// Fill your int-array
String hexString1 = sr.readLine();
String hexString2 = sr.readLine();

int[] intArray = new int[2];
intArray[0] = Integer.parseInt(hexString1, 16);
intArray[1] = Integer.parseInt(hexString2, 16);

// Print result (in dec and hex)
System.out.println(intArray[0] + " = " + Integer.toHexString(intArray[0]));
System.out.println(intArray[1] + " = " + Integer.toHexString(intArray[1]));

Output: 输出:

51966 = cafe
47806 = babe

I'm guessing you mean ascii hex? 我猜你是说ascii十六进制? In that case there isn't a trivial way, but it's not hard. 在这种情况下,没有平凡的方法,但这并不困难。

You need to know exactly how the strings are stored in order to parse them. 您需要确切地知道字符串的存储方式才能解析它们。

IF they are like this: 如果他们是这样的:

1203 4058 a92e

then you need to read the file in and use spaces and linefeeds (whitespace) as separators. 那么您需要读入文件,并使用空格和换行符(空格)作为分隔符。

If it's: 如果它是:

0x1203
0x4058

That's different yet 那不一样

and if it's: 并且如果是:

12034058...

That's something else. 那是另外一回事。

Figure out how to get it into strings where each string ONLY contains the hex digits of a single number then call 弄清楚如何将其分成字符串,其中每个字符串仅包含单个数字的十六进制数字,然后调用

Integer.parseInt(string, 16)

For strings that may have "0x" prefix, call Integer.decode(String) . 对于可能带有"0x"前缀的字符串 ,请调用Integer.decode(String) You can use it with Scanner , 您可以将其与Scanner一起使用,

try (Scanner s = new Scanner("0x11 0x22 0x33")) {
  while (s.hasNext()) {
    int num = Integer.decode(s.next());
    System.out.println(num);
  }
}
catch (Exception ex) {
  System.out.println(ex);
}

Unfortunately, unless the input is very short, Scanner is ridiculously slow. 不幸的是,除非输入非常短,否则Scanner的速度非常慢。 Here is an efficient hand-made parser for hexadecimal strings: 这是一个有效的手工解析器,用于十六进制字符串:

static boolean readArray(InputStream stream, int[] array) {

  int i = 0;

  final int SPACE = 0;
  final int X = 1;
  final int HEXNUM = 2;
  final int ERROR = -1;

  for (int next_char= -1, expected = SPACE; i <= array.length && stream.available() > 0 && expected != ERROR; next_char = stream.read()) {

    switch (expected)  {

      case SPACE:
        if (Character.isWhitespace(next_char))
          ;
        else if (next_char == '0') {
          array[i] = 0;
          expected = X;
        }
        else {
          LOGGER.e("unexpected '" + next_char + "' for " + expected + " at " + i);
          expected = ERROR;
        }
        break;

      case X:
        if (next_char == 'x' || next_char == 'X') {
          expected = HEXNUM;
        }
        else {
          LOGGER.e("unexpected '" + next_char + "' for " + expected + " at " + i);
          expected = ERROR;
        }
        break;

      case HEXNUM:
        if (Character.isDigit(next_char)) {
          array[i] *= 16;
          array[i] += next_char - '0';
        }
        else if (next_char >= 'a' && next_char <= 'f') {
          array[i] *= 16;
          array[i] += next_char - 'a' + 10;
        }
        else if (next_char >= 'A' && next_char <= 'F') {
          array[i] *= 16;
          array[i] += next_char - 'A' + 10;
        }
        else if (Character.isWhitespace(next_char)) {
          i++;
          expected = SPACE;
        }
        else {
          LOGGER.e("unexpected '" + next_char + "' for " + expected + " at " + i);
          expected = ERROR;
        }
    }
  }
}
if (expected == ERROR || i != array.length) {
  LOGGER.w("read " + i + " hexa integers when " + array.length + " were expected");
  return false;
}
return true;

Read in the values as strings, and call Integer.valueOf with a radix of 16. 读取值作为字符串,然后以16为基数调用Integer.valueOf。

See javadoc here: JavaSE6 Documentation: Integer.valueOf(String, int) 请参阅以下Javadoc: JavaSE6文档:Integer.valueOf(String,int)

A Scanner might be useful. Scanner可能会有用。 Are the numbers in your file prefixed with 0x ? 文件中的数字是否以0x If so you'll have to remove that and convert to an integer: 如果是这样,则必须将其删除并转换为整数:

// using StringReader for illustration; in reality you'd be reading from the file
String input = "0x11 0x22 0x33";
StringReader r = new StringReader(input);

Scanner s = new Scanner(r);
while (s.hasNext()) {
    String hexnum = s.next();
    int num = Integer.parseInt(hexnum.substring(2), 16);
    System.out.println(num);
}

If they're not prefixed with 0x it's even simpler: 如果它们不带0x前缀,则更为简单:

String input = "11 22 33";
StringReader r = new StringReader(input);

Scanner s = new Scanner(r);
while (s.hasNext()) {
    int num = s.nextInt(16);
    System.out.println(num);
}

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