Programming Studio #4ECE 190Programming Studio #4• Topics this week:• Systematic Decomposition• Memory Addressing Modes• In Studio Assignment– LC-3 Programming AssignmentAnnouncements• MP1– Due Wednesday, February 17that 5PM– Read handout carefully– Cheating Policy is on the website• Exam 1 – Thurs 2/25 – 7pm-9pm– Review Session: TBA– Report conflicts to professors NO LATER THAN Feb 17thSystematic Decomposition• Do not go straight to coding• Start with a problem in plain language – Break it into smaller steps progressively– Continue until steps small enough– Finally, steps can be turned into codeExample: Multiplication• Problem: – Design a multiplier program in LC-3 binary. The program should load numbers from x3100, x3101. The numbers are multiplied, and stored back in x3102.• Given algorithm: Add X to itself Y times. – X * Y = Z. – 5 x 3 = 5 + 5 + 5– Loop using branch instructions.First Phase: Large Steps• First few steps help show program flow • For multiplication we have 3 main steps• Why do we initialize registers?Middle Phase: Small Steps• Break large steps into smaller steps• Specify program flow more completely • Include multiplication algorithm• Add control flow to allow loop behaviorFinal Phase: Pseudo Code• Take small steps and convert to pseudo• Choose which registers hold which data– R2 stores running sum– R4 and R5 to hold numbers• Move from pseudo instructions into assembly instructions. • Why don't we need to initialize R4 or R5?ASM and Machine CodeAddressing Modes• Need to move data in/out of registers from/to memory• Several ways to specify the memory location– PC Relative– Base + Offset– Indirect– LEAPC Relative• Syntax:– LD Rx, imm_val | Rx <= MEM[PC1+imm_val] • Uses the incremented PC of the current instruction plus an immediate value to calculate address:– Address = PC + 1 + imm_val• imm_val is sign extended to 16 bits• Used when data is in a single location close to the instruction• Example:Address Contentsx3005 LD R1, x3x3006 ....x3007 ....x3008 ....x3009 x22• R1 receives contents of MEM[x3006+x3] = MEM[x3009] = x22Base + Offset• Syntax:– LDR Rx, Ry, imm_val | Rx <= MEM[Ry+imm_val] • Uses the contents of a register plus immediate value to calculate address:– Address = Ry + imm_val• Used when data is too far to access with PC Relative mode, or when accessing many values stored close to an address not known when writing the program• Example: (Assuming R2 holds the value x5000)Address Contentsx3005 LDR R1, R2, x0x3005 LDR R3, R2, x1… ....x5000 x1Ax5001 x22• R1 receives contents of MEM[x5000+x0] = x1A• R3 receives contents of MEM[?] = ?• NOT USED TO COPY CONTENTS FROM ONE REGISTER TO THE OTHER!!!Indirect• Syntax:– LDI Rx, imm_val | Rx <= MEM[MEM[PC1+imm_val]] • Uses the incremented PC of the current instruction plus an immediate value to calculate an address. Then reads the contents of that memory location and uses them as the address from where actual data is accessed:– Address = value at location [PC + 1 + imm_val]• imm_val is sign extended to 16 bits• Used when data is in a single location far from the instruction• Example:Address Contentsx3005 LDI R1, x1x3006 ....x3007 x4000… ....x4000 x22• R1 receives contents of MEM[MEM[x3006+x1]] = MEM[MEM[x3007]] = MEM[x4000] = x22Load Effective Address• Syntax:– LEA Rx, imm_val | Rx <= PC1+imm_val• Uses the incremented PC of the current instruction plus an immediate value to calculate address and places it in the register:– Address = PC + 1 + imm_val• imm_val is sign extended to 16 bits• Used when we need to load an address into a register (for use in LDR/STR)• Example:Address Contentsx3003 xBEx3004 xEFx3005 LEA R1, -3x3006 LDR R0, R1, x0x3007 LDR R2, R1, x1• R1 receives x3005+1-3 = x3003, R0 Receives?, R2 Receives?Programming Studio Assignment• Sum a list of N numbers and store result in x4000– Load N from address x3100– The list of numbers starts at x3101– Use systematic decomposition to break the problem into steps, then create the machine code.– You may assume that N >0.• Hint: Use LD/ST, LDR/STR, LDI/STI, LEA to your