emif配置如下：
当我执行：
*（*uint16 )(0x106001) = 0x1234;
时，
以下地址线io在写入时会按如下状态置吗？
| EM1A[13] | 0 |
| EM1A[12] | 1 |
| EM1A[11] | 1 |
| EM1A[10] | 0 |
| EM1A[9] | 0 |
| EM1A[8] | 0 |
| EM1A[7] | 0 |
| EM1A[6] | 0 |
| EM1A[5] | 0 |
| EM1A[4] | 0 |
| EM1A[3] | 0 |
| EM1A[2] | 0 |
| EM1A[1] | 0 |
| EM1A[0] | 0 |
| EMB1A[1] | 0 |
| EMB1A[0] | 1 |
数据会一次完成写入0x1234，还是会先往0x6001这个地址先后写入0x34和0x12？
另外，还有一个疑问，dsp与fpga通过emif异步方式数据交互时，在不进行数据读写时，emif外设相关引脚处于什么状态，比如地址线在不进行读写时，处于什么电平状态？
