INTA
- interrupt acknowledge, 80286 CPU响应INTR时与中断控制器之间的握手信号
Circuit of INTA
Tutorial 4
Question 1
(a)
- D0 activate -> D0 = 1, D1 = 0, D2 = 0 -> ISR 0
- D1 activate -> D1 = 0, D0 = 0, D2 = 0 -> ISR 1
- D2 activate -> D2 = 0, D1 = 0, D0 = 0 -> ISR 2
- 通过查找Figure 2得到
- ISR 0 -> A0 = 0, A1 = 1
- ISR 1 -> A0 = 1, A1 = 0
- ISR 2 -> A0 = 0, A1 = 0
- 根据Figure 1得到，A0和A1对应数据总线的低两位
- Interrupt Type Number: ISR0: 2H, ISR1: 1H, ISR2: 0H
- 最后通过base+offset(interrupt type number)计算ISR starting address的physical address(存疑)
(b)
- D2 has the highest priorty because we don’t care D0 and D1 when D2 = 1, 参考Figure 2
(c)
Process control
what is process
- program是静态代码，而process是程序的一次执行过程
- 操作系统(OS)将application转化为process, CPU上执行的任何一个task都必须是process
- process的核心要素
- program code
- set of data
- process control block
- process indetification: indentifiers包含process id/parent process id/user id
- process state information:
- user visible registers
- control and status register
- stack pointers
- scheduling and state information
- process control information
链接