/CSI2P-II-Mini1

Primary LanguageC

CSI2P-II-Mini1

Introduction

  • Let’s consider a CPU, which has eight 32 bits registers r0-r7 and a 256 byte memory.
  • In this project, you need to implement a calculator.The input is a list of expressions consisting of integers, operators(+,-,*,/,=), and three variables x,y, z; and the output is a list of assembly codes.
  • The instructions of the CPU are listed in the in the table below The time to execute each instruction (in clock cycles) is also listed.

Instruction Set Architecture(I)

Opcode Operand1 Operand2 Meaning Cycles
MOV Register1 Register2 Move data from register2 to register1 10
MOV Register1 Constant Set the value of register1 constant 10
MOV Register1 [Addr2] Move the data (4 bytes) in memory addressed Addr2 to register1. Note that Addr2 must be a multiple of 4. 200
MOV [Addr1] Register2 Move the data (4 bytes) from register2 to the memory addressed Addr1. Note that Addr1 must be a multiple of 4. 200

Instruction Set Architecture(II)

Opcode Operand1 Operand2 Meaning Cycles
ADD Register1 Register2 Add the values in register1 to register2 and store the result in register1 10
SUB Register1 Register2 Subtract the value in register2 from the value in register1 and store the result in register1. 10
MUL Register1 Register2 Multiply the values in register1 to register2 and store the result in register1 30
DIV Register1 Register2 Divide the value in register1 by the value in register2 and store the result in register1.Note it is the integer division. 50
EXIT Constant Stop the program with a constant signal,whose value is specified as follows. 0:exit normally 1:the expression cannot be evaluated 20

Variables

  • The initial value of variables, x, y, and z, are stored in memory [0], [4], and [8] respectively. You need to read those initial values first.

  • After the evaluation of the assembly code, the answer of the variables x, y, z needs to be stored in the registers r0, r1, and r2 respectively.

Example

  • Input: x = z + 5

  • Output :

    MOV r0, [8]

    MOV r1, 5

    ADD r0, r1

    MOV [0], r0

    MOV r0, [0]

    MOV r1, [4]

    MOV r1, [8]

    EXIT 0

Total clock cycles

  • Each instruction has an expected runtime, which is specified by clock cycles, as shown in the table.

  • The runtime of a program is the summation of the clock cycles of all instructions.

  • Example: the following code has 1040 clock cycles

    MOV r0, [8] 200 cc

    MOV r1, 5 10 cc

    ADD r0, r1 10 cc

    MOV [0], r0 200 cc

    MOV r0, [0] 200 cc

    MOV r1, [4] 200 cc

    MOV r1, [8] 200 cc

    EXIT 0 20 cc

Error handler

  • If the expression is illegal, such as

    • x = 5 +
    • y = 7 / 0
    • z = --2
    • ... and so on

  • You should consider all kinds of possibilities.

  • Your final output should be EXIT 1

Contest

  • The project has 2 parts:
    1. The 5 basic testcases will be provided by TAs
    2. The contest: There will be 20 testcases at demo time, each represents 5 points. Besides, the code with the less total clock cycles is better. Five top winners will get extra credits.

Submission / Demo

  • Submission Deadline: 4/20(五) 中午 12:00

    • 上傳至iLMS
    • 命名為 學號.c

  • Demo 時間: 4/20(五) 下午 1:20 ~ 3:30

    • 請準時出席課堂，我們會唱名請大家來臺前，會使用iLMS上傳的code來Demo
    • Demo完成後會公布過的測資數與cycle數
    • 不要使用別人的code，若抓到抄襲直接零分並依校規處置