CS355 Sylabus

Step 1 of the instruction execution cycle

OK, the CPU now starts the instruction execution cycle.
First, what is the state of the computer:
- MPC contains 0
- The PC register will contain the address of the memory that contains the next assembler instruction to be executed. (Note: when you first turn on the computer, PC will be reset to 0)
- IR does not contain any useful information
- The memory contains the program
The first step that the CPU must perform is to fetch the data at address given by the PC register from memory and store it in the IR register.

Recall the numbering of the registers:

Register	Name	Usage
R0	Program Counter (PC)	Holds address of next instruction
R1	Accumulator (AC)	Main working/scratch register
R2	Stack pointer (SP)	Points to top of stack
R3	Instruction Register (IR)	Holds currect instruction
R4	Temp Instruction Register (TIR)	Use to decode instruction
R5	0 (ZERO)	Hardwired Constant (like g0 in SPARC)
R6	+1 (ONE)	Hardwired Constant (like g0 in SPARC)
R7	-1 (MINUS ONE)	Hardwired Constant (like g0 in SPARC)
..	....	...

Recall the first few instructions of the mirco-program:

ADDR   Micro Assembler Instruction                      Comment
===========================================================================
 0:    mar := pc; rd;                                   {main loop}
 1:    pc := pc + 1; rd; mbr;                           {increment pc, MBR}
 2:    ir := mbr; if n then goto 31;                    {save, decode mbr}
 3:    tir := lshift(ir + ir); if n then goto 22;
 4:    tir := lshift(tir); if n then goto 13;           {000x or 001x?}
 5:    alu := tir; if n then goto 9;                    {0000 or 0001?}
       ....

OK, let's rock and roll:
1. MPC = 0, so the datapath fetches and executes mar := pc; rd;
  
  After running the 4-phase clock from phase 1 to phase 4, the datapath would have performed the following:
  - Address in the PC register is copied into MAR and sent out on the address bus
  - A READ operation is started
    - In other words: fetch the data at memory location given by PC !
  - Cond code = 00 will update MPC = 1 (0 + 1), and next micro-instruction fetched (and executed by the datapath) is at address 1.
2. Now MPC = 1, so the datapath fetches and executes pc := pc + 1; rd; mbr;
  
  After running the 4-phase clock from phase 1 to phase 4, the datapath would have performed the following:
  - Update PC = PC + 1 (PC register now points to the next assembler instruction in memory)
  - The READ operation is continued and during phase 3, the data on the databus will be copied into the MBR (because the mbr bit is set !)
    - In other words: the instruction memory location given by PC is now copied into the MBR
  - The cond code = 00 will update MPC = 2 (1 + 1), and next micro-instruction fetched (and executed by the datapath) is at address 2.
3. Now MPC = 2, so the datapath fetches and executes ir := mbr; if n then goto 31;
  
  After running the 4-phase clock from phase 1 to phase 4, the datapath would have performed the following:
  - Copy the data from MBR into the IR register
    - In other words: the instruction memory location given by PC is now copied into the IR register
    - The instruction at PC has been fetched !!!
  - The ALU's output is equal to the code in the assembler instruction and the N-flag is equal to the left most bit in the assembler instruction.
  - The cond code = 01 and the value of the ALU's N bit will:
    - update MPC = 3 (2 + 1) if the left most bit of the assembler instruction = 0.
    - update MPC = 31 (branch addr) if the left most bit of the assembler instruction = 1.
Conclusion:
DEMO: (Demo instruction fetch)
- Run: /home/cs355000/bin/cs355-demo-computer
Do the following:
1. Press I, then press r
2. Maximize the window
3. Toggle key 1 TWICE - this resets the computer
4. Toggle key 1 REPEATED and you will see this:
  - CPU sends address 0 to memory
  - Memory returns the instruction at address 0 (which is: 0000000000001010)
  - This instruction is store in IR (R3 - the 4th register)