"The two BD entries allows for an EVEN BD and ODD BD entry for each endpoint direction.This allows the microprocessor to process one BD while the USB-FS is processing the other BD"这句话要如何理解咧！为什么BD提供两个entry就能做到在USB-FS同时处理两个BD呢？
在RM里面，你可以在"The two BD entries allows for an EVEN BD and ODD BD entry for each endpoint direction.This allows the microprocessor to process one BD while the USB-FS is processing the other BD"前面看到“The BDT resides on a 512 byte boundary in system memory and is pointed to by the BDT Page Registers. Every endpoint direction
requires two eight-byte Buffer Descriptor entries. Therefore, a system with 16 fully bidirectional endpoints would require 512 bytes of system memory to implement the BDT.”结合起来就好理解了，一个BD占用8字节，每个endpoint（IN or OUT）占用两个BD,USB模块最多用到16个endpoint,因此整个BDT占用了16*2*16=512个字节,所以这个BDT的起始地址要放在512字节对齐的地方，因此BDT的结构可理解为：
BD0//EP0 IN
BD1//EP0 IN
BD2//EP0 OUT
BD3//EP0 OUT
...
BD30//EP15 OUT
BD31//EP15 OUT
所以EVEN BD是指BD0,BD2,...,BD30, ODD BD是指BD1.BD3,...,BD31,当应用程序要发送数据是，比方说EP0 IN方向，USB device模式，发送信息给HOST(PC),那么将用到BD0和BD1，MCU初始化BD0后，把控制权交给USB-FS引擎处理发送，然后不用等待其发送完毕，就可以直接初始化BD1，为下一次发送做好准备，从而达到“Double buffering BDs in this way allows the USB-FS to easily transfer data at the maximum throughput provided by USB.”这样的效果。
我也问几个BDT相关的问题：
1，BDT里面的32bit控制位部分（CtrlReg），25:16是BC，即Byte Count，RM的说明是：
The Byte Count bits represent the 10-bit Byte Count. The USB-FS SIE changes this field upon the completion of a RX transfer with the byte count of the data received.
请问，TX操作也需要BC吗？
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