英文:
Getting Date Time in Unix Time as Byte Array which size is 8 bytes with Java
问题
我知道我可以通过4个字节来获取它，就像这样：
int unixTime = (int)(System.currentTimeMillis() / 1000);
byte[] productionDate = new byte[]{
(byte) (unixTime >> 24),
(byte) (unixTime >> 16),
(byte) (unixTime >> 8),
(byte) unixTime
};
但是是否有一种使用移位来获取8个字节的方法呢？
英文:
I know I can get it in 4 bytes like so:
int unixTime = (int)(System.currentTimeMillis() / 1000);
byte[] productionDate = new byte[]{
(byte) (unixTime >> 24),
(byte) (unixTime >> 16),
(byte) (unixTime >> 8),
(byte) unixTime
};
but is there a way to get it in 8 bytes using shifts?
答案1
得分: 4
当然，只需将其解读为有符号的long。
long unixTime = System.currentTimeMillis() / 1000L;
byte[] bytes = new byte[] {
(byte) (unixTime >> 56),
(byte) (unixTime >> 48),
(byte) (unixTime >> 40),
(byte) (unixTime >> 32),
(byte) (unixTime >> 24),
(byte) (unixTime >> 16),
(byte) (unixTime >> 8),
(byte) unixTime
};
或者，使用 NIO 的 ByteBuffer：
ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES)
.putLong(unixTime);
byte[] bytes = buffer.array();
英文:
Sure, just read as a signed long.
long unixTime = System.currentTimeMillis() / 1000L;
byte[] bytes = new byte[] {
(byte) (unixTime >> 56),
(byte) (unixTime >> 48),
(byte) (unixTime >> 40),
(byte) (unixTime >> 32),
(byte) (unixTime >> 24),
(byte) (unixTime >> 16),
(byte) (unixTime >> 8),
(byte) unixTime
};
Alternatively, using NIO ByteBuffer
ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES)
.putLong(unixTime);
byte[] bytes = buffer.array();
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