Solutions for Homework 2 in Computer Architecture I Course, Lecture notes of Architecture

Download Solutions for Homework 2 in Computer Architecture I Course and more Lecture notes Architecture in PDF only on Docsity! CSSE 232 – Computer Architecture I Rose-Hulman Institute of Technology Winter 2003-2004 Computer Science and Software Engineering Prof. Archana Chidanandan Homework 2 - Solutions (Assembly Language and Machine Language) Maximum points : 40 points Directions This assignment is due Friday, Dec. 19th for Sections 1 and 2. Submit your solutions on a separate sheet of paper. Learning Objectives In the process of completing this homework assignment, students will develop their abilities to • Translate assembly language instructions into machine language • Interpret a sequence of bits and determine what it represents. • Determine addressing in branches and jumps. • Interpret instruction formats and fields. • Implement algorithms procedural abstraction in assembly language. Problems 1. [6 points] In Homework 1, you wrote a program that could calculate the product of two numbers. Re-write the program, so that the main program reads the two values from memory, uses procedure “Product” to determine the product of the two numbers, and writes the product into a location in memory. You must also write the procedure “Product”. Again, you may NOT use the MIPS mult, multu, mul, mulo or muluo instructions. You MUST use a loop structure. The program must follow the MIPS register conventions (page 138, Figure 3.11). .text .globl main main: la $s0, A lw $s0, 0($s0) # Read the value of “A” from memory la $t1, B lw $s1, 0($t1) # Read the value of “B” from memory. move $a0, $s0 # Copy the arguments "A" and "B" into the # argument registers move $a1, $s1 jal Product # Call procedure "Product" move $t4, $v0 # Store return value "c" in temporary # register. la $t3, C sw $t4, 0($t3) # Store the value of “c” in memory li $v0, 10 syscall # Ready to quit CSSE 232 – Computer Architecture I Rose-Hulman Institute of Technology Winter 2003-2004 Computer Science and Software Engineering Prof. Archana Chidanandan Product: # Procedure to determine the product move $t0, $a0 # Move the paraments into temporary # registers. move $t1, $a1 li $t3, 0 # i = 0 (to keep track of the value “b” li $t4, 0 # c = 0 (initialize the result to 0) loop: beq $t3, $t1, done # while (i is not equal to b ) # continue adding a to c. add $t4, $t4, $t0 # c= c + a; addi $t3, $t3, 1 # i = i + 1 j loop done: move $v0, $t4 # Move return value "c" to the return # value register. jr $ra # Return to the calling procedure .data A: .word 5 B: .word 20 C: .word 0 2. [8 points, 4 each] In the in-class exercise, we looked at a procedure “Move” (p04- 2.asm), that uses a position component and a velocity component to determine a new position which lies within the specified boundaries. The procedure (the name has been changed to “MoveObject”) is provided in the file, hmwk02-2.asm. a) Write a procedure “MoveBall” that calls the “MoveObject” procedure to update the “x” and “y” positions of the ball. b) Write a main program that initializes the x and y positions and velocities of the ball and then calls the procedure “MoveBall” repeatedly to update the “x” and “y” positions. The program must follow the MIPS register conventions (page 138, Figure 3.11). You may not modify the “MoveObject” procedure. .text # Text section of the program (as opposed to # data). .globl main .globl MoveBall main: li $s0, 2 # Assign the x-pos li $s1, 3 # Assign the x-vel li $s2, 5 # Assign the y-pos li $s3, 7 # Assign the y-vel loop: move $a0, $s0 # Move x-pos to $a0 move $a1, $s1 # Move x-vel to $a1 move $a2, $s2 # Move y-pos to $a2 move $a3, $s3 # Move y-vel to $a3 CSSE 232 – Computer Architecture I Rose-Hulman Institute of Technology Winter 2003-2004 Computer Science and Software Engineering Prof. Archana Chidanandan li $a2, -1 li $a3, -1 li $v1, -1 jr $ra .data # Data section of the program RA: .word 0 SOne: .word 0 STwo: .word 0 SThree: .word 0 3. [6 pts, 2 each] Bits have no inherent meaning. However, given the rules to interpret them, a sequence of bits can represent different values and have meaning. Given the bit pattern, 0x8dad0000, what does it represent, assuming that it is a) an unsigned integer? b) a 2’s complement number? c) a MIPS instruction? Hint: Use a calculator where appropriate. a) 0x8dad000 = (1000 1101 1010 1101 0000 0000 0000 0000)2 = (2376925184)10 b) 2’s complement number. The sign-bit (MSB) is “1”, therefore this is the 2’s complement representation of a negative number. X = 1000 1101 1010 1101 0000 0000 0000 0000 X + 1 = 0111 0010 0101 0011 0000 0000 0000 0000 = 1918042112 Therefore, 0x8dad000 = -1918042112 d) MIPS instruction 100011 01101 01101 0000 0000 0000 0000 lw $13 $13 0x0000 Therefore, 0x8dad0000 represents the following MIPS instruction: lw $t5, 0($t5) 4. [9 pts, 3 each] For the following MIPS assembly language instructions, list the machine language fields and their binary and hexadecimal values: a) ori $2, $0, 10 b) jal main # main is at address 0x00400020 and the current instruction # is in the same page as “main” c) add $t7, $t2, $0 CSSE 232 – Computer Architecture I Rose-Hulman Institute of Technology Winter 2003-2004 Computer Science and Software Engineering Prof. Archana Chidanandan a) ori $2, $0, 10 This is an I-type instruction. ori $0 $2 0x000A 001101 00000 00010 0000 0000 0000 1010 Group into 4 bits and obtain the hex representation. 0x340200a b) jal main # main is at address 0x00400020 and the current instruction is in the same page as “main”. This is a J-type instruction. jal 0x00400020 000011 0000 0000 0100 0000 0000 0000 0010 0000 0000 1100 0001 0000 0000 0000 0000 1000 Hex representation: 0x0c100008 c) add $t7, $t2, $0 This is an R-type instruction add $t2 $0 $t7 000000 01010 00000 01111 00000 100000 = 0x01407820 5. [4 pts] Assume that the MIPS instruction beq $t0, $s0, Label is located at address 0x0400 5678, and that Label is located at address 0x0400 1234. What will the binary value of the address field be? Hint: Remember that the offset is relative to the instruction following the branch, and that all branch targets must be word aligned. “beq” in MIPS uses PC-relative addressing for the address field a) Determine the address in PC: 0x400 5678 + 4 = 0x4000 567c b) Determine the number of addresses between address in the PC and the target instruction address: 0x4000 567c – 0x4000 1234 = 4448 addresses ot bytes. c) Determine number of instructions: 4448/4 = 0x1112 instructions. Since, we have to go back, we must obtain the 2’s complement value. 0001 0001 0001 0010 => 1110 1110 1110 1110 => 0xEEEE Therefore, the 16-bit immediate value in the “beq” instruction will be 0xEEEE. 6. [4 pts] Assume that the MIPS instruction j Label is located at address 0x0400 5678, and that Label is located at address 0x0400 1234. What will the binary value of the target address field be? Hint: Remember that all branch targets must be word aligned. “j” instructions uses Pseudo-direct addressing. The 32-bit address needs to be fit into a 26-bit field. a) Chop off the last two bits, as they are always “00” for any instruction address in MIPS. b) Since the two instructions are on the same page (observe the most significant 4 bits, they are equal), the most significant 4 bits need not encoded in the label field. 0000 0100 0000 0000 0001 0010 0011 0100 = 0x100048D CSSE 232 – Computer Architecture I Rose-Hulman Institute of Technology Winter 2003-2004 Computer Science and Software Engineering Prof. Archana Chidanandan 7. [3 pts] What sequence of MIPS instructions starting at address 0x0400 5678 could be used to branch to address 0x4400 1234? The two addresses are not on the same page(the 4 MSBs are not the same for the addresses of the two instructions). Therefore, the “j” instruction cannot be used. However, if the 32-bit value can be copied into a register, the “jr” instruction could be used. li $at, 0x44001234 jr $at