我正在嘗試從字節數組中提取 6 個字節并將它們轉換為 48 位有符號整數值（即 Java 長整數）。如何做到這一點？編輯：例如，如果字節數組具有以下之一：byte[] minInt48 = new byte[] { (byte)0x80, 0, 0, 0, 0, 0 };
byte[] maxInt48 = new byte[] { (byte)0x7F, (byte)0xFF, (byte)0xFF, (byte)0xFF, (byte)0xFF, (byte)0xFF };如何將其解析為 Java 原語（即 Java long）以保留符號值？
1 回答
慕尼黑8549860
TA貢獻1818條經驗 獲得超11個贊
首先，您需要了解符號擴展。您需要將每個字節視為無符號值。
long v = 0;
byte q = -2; // unsigned value of 254
v = v + q;
System.out.println(v); // prints -2 which is not what you want.
v = 0;
v = v + (q & 0xFF); // mask off sign extension from q.
System.out.println(v); // prints 254 which is correct.
這是一種方法。
long val = 0;
byte[] bytes = { -1, 12, 99, -121, -3, 123
};
for (int i = 0; i < bytes.length; i++) {
// shift the val left by 8 bits first.
// then add b. You need to mask it with 0xFF to
// eliminate sign extension to a long which will
// result in an incorrect conversion.
val = (val << 8) | ((i == 0 ? bytes[i]
: bytes[i] & 0xFF));
}
System.out.println(val);
添加回答
舉報
0/150
提交
取消