CS 520 Programming Project 1: Simulator for APEX with in-order issue solved

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Description

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PROJECT DESCRIPTION PART A:

This project requires you to implement a cycle-by-cycle simulator for an in-order APEX pipeline with 5 pipeline stages, as shown below: The EX stage is NOT pipelined and has a latency of 1 cycle and implements all operations that involve integer arithmetic (ADD, SUB, ADDC, address computation of LOADs and STOREs etc.), excepting for the MUL (for multiply) instruction.

MUL, which is a register-to-register instruction like the ADD, uses the non-pipelined EX stage, to implement the multiplication operation but spends two cycles in it. Once the MUL instruction starts, it ties up the EX stage for two cycles.

Assume for simplicity that the two values to be multiplied have a product that fits into a single register. Instruction issues in this pipeline take place in program-order and a simple interlocking logic is used to handle dependencies. For PART A of this project, this pipeline has no forwarding mechanism.

Registers and Memory:

Assume that there are 16 architectural registers, R0 through R15. The code to be simulated is stored in a text file with one ASCII string representing an instruction (in the symbolic form, such as ADD R1, R4, R6) in each line of the file. Memory for data is viewed as a linear array of integer values (4 Bytes wide).

Data memory ranges from 0 through 3999 and memory addresses correspond to a Byte address that begins the first Byte of the 4-Byte group that makes up a 4 Byte data item. Instructions are also 4 Bytes wide. Registers are also 4 Bytes wide.

Fetch D/RF EX MEM WB Two cycle delay for MUL, one cycle delay for all other instructions 2 of 3

Instruction Set:

The instructions supported are:  Register-to-register instructions: ADD, SUB, MOVC, AND, OR, EX-OR and MUL.

As stated earlier, you can assume that the result of multiplying two registers will fit into a single register.  MOVC , moves literal value into specified register. The MOVC uses the ALU stages to add 0 to the literal and updates the destination register from the WB stage.  Memory instructions: LOAD, STORE – both include a literal value whose content is added to a register to compute the memory address.  Control flow instructions: BZ, BNZ, JUMP and HALT.

Instructions following a BZ, BNZ and JUMP instruction in the pipeline should be flushed on a taken branch. The zero flag (Z) is set only by arithmetic instructions when they are in the WB stage. The dependency that the BZ or BNZ instruction has with the immediately prior arithmetic instruction (ADD, MUL, SUB) that sets the Z flag has to be implemented correctly. The semantics of BZ, BNZ, JUMP and HALT instructions are as follows:  The BZ instruction cause in a control transfer to the address specified using PC-relative addressing if the Z flag is set.

The decision to take a branch is made in the Integer ALU stage. BNZ is similar but causes branching if the Z flag is not set. BZ and BNZ target 4-Byte boundaries (see example below).  JUMP specifies a register and a literal and transfers control to the address obtained by adding the contents of the register to the literal. The decision to take a branch is made in the Integer ALU stage.

JUMP also targets a 4-Byte boundary.  The HALT instruction stops execution as soon as it is decoded but allows all prior instructions in the pipeline to complete. The instruction memory starts at Byte address 4000. You need to handle target addresses of JUMP correctly – what these instructions compute is a memory address. However, all your instructions are stored as ASCII strings, one instruction per line in a SINGLE text file and there is no concept of an instruction memory that can be directly accessed using a computed address.

To get the instruction at the target of a BZ, BNZ or JUMP, a fixed mapping is defined between an instruction address and a line number in the text file containing ALL instructions:  Physical Line 1 (the very first line) in the text file contains a 4 Byte instruction that is addressed with the Byte address 4000 and occupies Bytes 4000, 4001, 4002, 4003.  Physical Line 2 in the text file contains a 4 Byte instruction that is addressed with the Byte address 4004 and occupies Bytes 4004, 4005, 4006, 4007.  Physical Line 3 in the text file contains a 4 Byte instruction that is addressed with the Byte address 4008 and occupies Bytes 4008, 4009, 4010, 4011 etc.

The targets of all control flow instructions thus have to target a 4_byte boundary. So when you simulate a JUMP instruction whose computed target has the address 4012, you are jumping to the instruction at physical line 4 in the text file for the code to be simulated. Register contents and literals used for computing the target of a branch should therefore target one of the lines in the text file.

Your text input file should also be designed to have instructions at the target to start on the appropriate line in the text file. Instructions are stored in the following format in the text file, one per line: 3 of 3 etc. Where arguments are registers or literals. Registers are specified using two or three characters (for example, R5 or R14). Literal operands, if any, appear at the end, preceded by an optional negative sign.

Simulator Commands:

Your simulator is invoked by specifying the name of the executable file for the simulator and the name of the ASCII file that contains the code to be simulated. Your simulator should have a command interface that allows users to execute the following commands: Initialize: Initializes the simulator state, sets the PC of the fetch stage to point to the first instruction in the ASCII code file, which is assumed to be at address 4000. Each instruction takes 4 bytes of space, so the next instruction is at address 4001, as memory words are 4 Bytes long, just like the integer data items. Simulate : simulates the number of cycles specified as and waits.

Simulation can stop earlier if a HALT instruction is encountered and when the HALT instruction is in the WB stage. Display: Displays the contents of each stage in the pipeline, all registers, the flag register and the contents of the first 100 memory locations containing data, starting with address 0. A skeleton code in C for this simulator that simulates only two instructions is included in a separate file. This simulator has most of the major data structures that you will need to use and also implements the simulated memory for instructions and data. You can extend this code by modifying it appropriately for this assignment.

PART B:

For the simulated pipeline, add the complete set of necessary forwarding logic to forward register values. Please document your code appropriately and check it out by using test code sequences. For PART A, it may be easier to add one instruction at a time and then test the simulator. You can write simple test code sequences yourself.

For PART B, a similar approach may be used by adding code to forward data between one specific pair of stages at a time. Again, use you own test code sequence.