[英]Java reading hex values into an array of type int
我有一個文件,其中包含以十六進制IS表示的整數,可以通過任何方式將所有這些數字存儲到整數數組中。
我知道你可以說int i = 0x
但是在讀取出現錯誤的值時我不能這樣做?
提前致謝!
您可能想要遍歷Integer.parseInt(yourHexValue, 16)
。
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]));
輸出:
51966 = cafe
47806 = babe
我猜你是說ascii十六進制? 在這種情況下,沒有平凡的方法,但這並不困難。
您需要確切地知道字符串的存儲方式才能解析它們。
如果他們是這樣的:
1203 4058 a92e
那么您需要讀入文件,並使用空格和換行符(空格)作為分隔符。
如果它是:
0x1203
0x4058
那不一樣
並且如果是:
12034058...
那是另外一回事。
弄清楚如何將其分成字符串,其中每個字符串僅包含單個數字的十六進制數字,然后調用
Integer.parseInt(string, 16)
對於可能帶有"0x"
前綴的字符串 ,請調用Integer.decode(String) 。 您可以將其與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);
}
不幸的是,除非輸入非常短,否則Scanner的速度非常慢。 這是一個有效的手工解析器,用於十六進制字符串:
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;
讀取值作為字符串,然后以16為基數調用Integer.valueOf。
請參閱以下Javadoc: JavaSE6文檔:Integer.valueOf(String,int)
Scanner
可能會有用。 文件中的數字是否以0x
? 如果是這樣,則必須將其刪除並轉換為整數:
// 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);
}
如果它們不帶0x
前綴,則更為簡單:
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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