英文:
Convert string of hex-encoded integers to int array in Java
问题
使用Java正确将十六进制字符串转换为整数的方法如下：
String s = "c4d2"; // 十六进制字符串
int n = Integer.parseInt(s, 16); // 使用十六进制解析
System.out.println(s + " -> " + n);
这会将十六进制字符串 "c4d2" 转换为整数 53956，与你所期望的结果一致。
请注意，与Python不同，Java的整数解析方法要求输入的字符串必须以正确的十六进制格式表示，并且不需要额外的字节顺序或二进制转换步骤。
英文:
For example, we have a string "c4d2", known it is a little-endian coded 16-bit int 53956. (It is NOT about conversion "c4d2" to int[] {0xc4, 0xd2}, but about coversion of "c4d2" to integer 53956). How to convert the string to int use Java?
Use python I do
s = b'\xc4\xd2'
n = int.from_bytes(s, "little")
print(f"{s} -> {n}")
output is right
b'\xc4\xd2' -> 53956
Use Java I do
String s = "c4d2"; // 53956
byte[] b = HexFormat.of().parseHex(s);
int n = ByteBuffer.wrap(b).order(ByteOrder.LITTLE_ENDIAN).getShort();
System.out.println(s + " -> " + Arrays.toString(b) + " -> " + n);
and got
c4d2 -> [-60, -46] -> -11580
that is wrong. But if I do conversion through binary string as
String s2 = Integer.toBinaryString(0xd2) + Integer.toBinaryString(0xc4);
int n2 = Integer.parseInt(s2, 2);
System.out.println(s + " -> " + s2 + " -> " + n2);
I got
c4d2 -> 1101001011000100 -> 53956
the result is right, but the method seems stange.
How to properly convert hex-string to int?
答案1
得分: 1
Java的short是有符号的，因此其最大值是32767，你发生了溢出。你可以使用Short.toUnsignedInt()：
String s = "c4d2"; // 53956
byte[] b = HexFormat.of().parseHex(s);
int n = Short.toUnsignedInt(ByteBuffer.wrap(b).order(ByteOrder.LITTLE_ENDIAN).getShort());
System.out.println(s + " -> " + Arrays.toString(b) + " -> " + n);
输出结果为：
c4d2 -> [-60, -46] -> 53956
英文:
Java's short is signed, so its max value is 32767 and you got overflow. You can use Short.toUnsignedInt():
String s = "c4d2"; // 53956
byte[] b = HexFormat.of().parseHex(s);
int n = Short.toUnsignedInt(ByteBuffer.wrap(b).order(ByteOrder.LITTLE_ENDIAN).getShort());
System.out.println(s + " -> " + Arrays.toString(b) + " -> " + n);
prints
> c4d2 -> [-60, -46] -> 53956
答案2
得分: 1
一种方法是反转这些值，然后使用Integer#parseInt方法，并指定基数为16。
String string = "c4d2";
StringBuilder reversed = new StringBuilder();
for (int index = string.length() - 1; index >= 0; index -= 2)
reversed.append(string, index - 1, index + 1);
int number = Integer.parseInt(reversed.toString(), 16);
输出，对于_number_。
53956
英文:
One approach would be to reverse the values, and then use the Integer#parseInt method, specifying a 16 for the radix.
String string = "c4d2";
StringBuilder reversed = new StringBuilder();
for (int index = string.length() - 1; index >= 0; index -= 2)
reversed.append(string, index - 1, index + 1);
int number = Integer.parseInt(reversed.toString(), 16);
Output, for number.
53956
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