1) Part a) As you know the memory address space of the LC-3 is 16 bits. If the MAR is loaded with the value xFE00, how does the hardware know to access the Keyboard Status Register or memory location xFE00?
There is no memory location xFE00, so the address control logic tells the hardware that it is accessing a device register.
Part b) The following assembly program is assembled, and run on the LC-3 Simulator.
.ORIG x3000 LD R0, ASCII0 TRAP x23 TRAP x23 TRAP x23 TRAP x23 TRAP x23 TRAP x25 ASCII0 .FILL x30 .END
Before it is executed, you set a breakpoint at x3003. What happens?
The number "0" is output twice, then the process stops.
Part c) At the end of an LD instruction midway through the execution of an LC-3 program, the contents of the condition codes are set as follows: N=1, Z=1, P=0. What can you infer?
This is impossible, so the machine is broken. Note: there is no instruction that can cause this to happen.
Part d) There are three addressing modes available to the assembly language programmer who wishes to load a value from memory into R5. If the load instruction is in a loop and each time through the loop, the next consecutive memory location is loaded into R5, which addressing mode is most appropriate to use? Explain why.
Base Register + offset (LDR), just increment the register holding the address in each iteration.
2) What is the output of the following program?
.ORIG x3000 LD R1, LETA LEA R0, BUFFER STR R1, R0, #0 ADD R1, R1, #1 STR R1, R0, #1 ADD R1, R1, #1 STR R1, R0, #2 ADD R1, R1, #1 STR R1, R0, #3 TRAP x22 LD R0, LF TRAP x21 LEA R0, STRING TRAP x22 TRAP x25 LF .FILL x000A LETA .FILL x0041 BUFFER .BLKW #4 STRING .STRINGZ "EFGH" .END
3) Part a) Reverse-assemble the binary program (convert the binary program into an assembly language program). Most of the instructions have already been reverse-assembled for you, so your job is to complete the task.
.ORIG x3000 0011000000000000 AND R0, R0, x0 0101000000100000 AND R1, R1, x0 0101001001100000 ADD R1, R1, x9 0001001001101001 BRn EE 0000100000000100 LD R2, FF 0010010000001000 LEA R3, FF STR R1, R3, #2 0111001011000010 LEA R7, DD 1110111000000011 EE NOT R5, R5 1001101101111111 BRnz DD 0000110000000001 NOT R4, R3 1001100011111111 DD LDR R6, R2, #1 0110110010000001 TRAP x25 1111000000100101 FF .FILL xFF00 1101000000000000 .FILL xFF00 1111111100000000 .FILL xFAFA 1111101011111010 .END
Part b) Generate the symbol table that an LC-3 assembler would create while assembling this program. You may not need all of the spaces provided.
Symbol Address EE x3008 DD x300B FF x300D
4) The input to the logic circuit is the 16 bits of the MAR. What information does the output provide?
The output is a 1 if the address in the MAR points to a device register (KBSR, KBDR, DDR, DSR). Note: This is performed by the address control logic.
5) The following program has been assembled and loaded into the LC-3 simulator. A breakpoint has been set on the TRAP x25 instruction. Your job is to trace the execution of the program until the breakpoint is reached. By trace we mean record the value of the PC at the beginning of each instruction in the order that they are executed. The first few have been done for you. You may not need all of the spaces provided.
Note: The instruction labeled A is at location x0100 in memory, and the instruction labeled START is at location x3000 in memory.
.ORIG x0000 .FILL x0100 .FILL x0101 .FILL x0102 .FILL x0103 .FILL x0104 .FILL x0105 .FILL x0106 .BLKW x00F9 A RET ; LOCATION x0100 RET RET RET RET RET RET .BLKW x2EF9 START AND R0, R0, #0 ; LOCATION x3000 BRz L1 LD R0, DATA L1 NOT R0, R0 BRn L2 TRAP x05 L2 TRAP x06 AND R0, R0, #0 TRAP x25 DATA .FILL x4040
PC Trace x3000 x3001 x3003 x3004 x3006 x0106 x3007 x3008
7) What does the following program do?
.ORIG x3000 AND R5, R5, #0 AND R3, R3, #0 ADD R3, R3, #8 LEA R0, BB LDR R1, R0, #1 LDR R1, R1, #0 ADD R2, R1, #0 AGAIN ADD R2, R2, R2 ADD R3, R3, #-1 BRp AGAIN LDR R4, R0, #0 AND R1, R1, R4 NOT R1, R1 ADD R1, R1, #1 ADD R2, R2, R1 BRnp NO ADD R5, R5, #1 NO TRAP x25 BB .FILL xFF00 .FILL x4000 .END
It puts 1 in R5 if the high byte of the data in location x4000 is equal to the low byte of the data in location x4000.
7) The table below shows a snapshot of the Program Counter, the 8 registers, and the condition codes (CC) of the LC-3 at six different times during the execution of a program: before the program executes, after the execution of instruction 1, after the execution of instruction 2, after the execution of instruction 3, after the execution of instruction 4, and after the execution of instruction 5. Fill in the missing values in the table as well as the missing parts of instructions 1, 4, and 5.
Initial After 1st After 2nd After 3rd After 4th After 5th CC Z N P P N P PC x3300 x3301 x3302 x3303 x3304 x3305 R0 x0000 x0000 x0000 x0000 x0000 x0000 R1 x1111 x1111 x1111 x1111 x1111 x1111 R2 x2222 x2222 x2222 x2222 x2222 x2222 R3 x3333 x3333 x3308 x3308 x3308 x3308 R4 x4444 x4444 x4444 x3303 x9FFF x9FFF R5 x5555 x5555 x5555 x5555 x5555 x5555 R6 x6666 x6666 x6666 x6666 x6666 x6666 R7 xFEFE xFEF0 xFEF0 xFEF0 xFEF0 x010F Instruction 1: AND R7, R7, #-16 Instruction 2: LEA R3, #6 Instruction 3: LEA R4, #0 Instruction 4: LD R4, #0 Instruction 5: NOT R7, R7