Question

此代码测试到底是什么？这是一个ID3标记库。我刚刚开始学习“和”。据我所知，“＆amp; 0x0F”将位移到右边。

非常感谢您对此if语句提供的任何说明。

protected static final int FLAGS_OFFSET = 5;
if ((bytes[FLAGS_OFFSET] & 0x0F) != 0) throw new UnsupportedTagException("Unrecognised bits in  header");

第5个字节包含一个标志：

ID3v2 flags             %abc00000

"%x is used to indicate a bit with unknown content."

a - Unsynchronisation Bit 7 in the 'ID3v2 flags' indicates whether or not unsynchronisation is used (see section 5 for details); a set bit indicates usage.

b - Extended header The second bit (bit 6) indicates whether or not the header is followed by an extended header. The extended header is described in section 3.2.

c - Experimental indicator The third bit (bit 5) should be used as an 'experimental indicator'. This flag should always be set when the tag is in an experimental stage.

Answer 1

如果两个操作数在该位中都设置为1，则&的结果中的位为1，例如：

  01010101
& 00001111
==========
  00000101

0x0F是二进制的1111。（这被称为＆＃34;掩码＆＃34;。）所以表达式说＆＃34;如果在索引5的字节中设置了最低4位中的任何一个，则抛出异常＆＃34;。

if（（bytes [FLAGS_OFFSET]＆amp; 0x0F）！= 0）抛出新的

1 个答案: