Homework 3_ A simple processor

The following instruction set is supported by a simple processor, which is similar to what we discussed in the class, with a few new instructions added. The format of most instructions is defined as follows.
bits 15:14 13:10 9 8:6 5:3 2:0 field unused opcode w src1 src2 dst
where the fields are defined as follows. opcode : operation to be performed by the processor w: write back ALU output to register file (1 = yes, 0 = no) src1: address of the first ALU operand in the register file src2: address of the second ALU operand in the register file dst: address in the register file where the output is written
For opcodes BEQ, BLEZ and JUMP, the 6 least significant bits (5:0) give an address in the instruction memory, which is byte-addressed. The opcode HALT has all operand bits (9:0) being 0. When an instruction has only two operands, the field for the unused operand is filled with 0-bits. For example, bits (5:3) for SLL are all zero because src2 is not used.
The opcode and meaning of these instructions are listed in the following table.
opcode Binary encoding Operation ADD 0x0 R[src1] + R[src2]  R[dst] SUB 0x1 R[src1] – R[src2]  R[dst] SLL 0x2 R[src1] << 1  R[dst] SRL 0x3 R[src1] 1  R[dst] INV 0x4 ~R[src1]  R[dst] XOR 0x5 R[src1] ^ R[src2]  R[dst] OR 0x6 R[src1] | R[src2]  R[dst] AND 0x7 R[src1] & R[src2]  R[dst] INCR 0x8 R[src1] + 1  R[dst] BRZ 0x9 If R[src1] = 0, branch to BAddr BLEZ 0xA If R[src1] ≤ 0, branch to BAddr
JUMP 0xE Jump to JAddr HALT 0xF Stop execution
1. In this part you are asked to recognize what a given segment of machine code does, by translating it into assembly code, and annotating it with expressions in a high level language like C. This process,
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from binary code back to source code, is called “disassembling.” Disassemble the following machine code into operations and arguments, e.g., ADD R1, R2, R3. Explain what the whole program does, either in plain English (e.g., it calculates the product of R1 and R2) or in C type of pseudo code (e.g, R3 = R1 * R2.) Note: you need to make up unique labels for your assembly code.
Addr: Code ----------------------- 0x00: 07 FF 0x02: 06 08 0x04: 28 08 0x06: 38 02 0x08: 02 47 0x0A: 3C 00
2. If the register files (8-bit, byte addressed) have the initial image as the following, what are the values of these 8 registers when the execution of the program in part 1 stops?
R0: 0001 1101 R1: 0000 0111 R2: 0000 0000 R3: 0000 0000 R4: 0000 0000 R5: 0000 0000 R6: 0000 0000 R7: 0001 0000
3. Translate the following segment of C program into assembly code for the machine given above. Note: you may assume two positive integers a and b are already given in register R0 and R1. By convention, put the returned value in R7.
while (a!=b) {
if (a b) { a = a - b; } else { b = b – a; }
} return a;