University of California at Berkeley
College of Engineering
Department of Electrical Engineering and Computer Science

CS61C, Spring 2008

## ...HW 4!

[TA] Eric - and the Elders Of 61C

Due Wednesday, Feb 24, 2010 @ 11:59pm

## Goals

This assignment will give you more practice with MIPS and make sure you're solid on MIPS procedure calling. By the end you should have no qualms about jaling all over the place. We'll also give you a bit of exposure to converting the squishy MIPS instructions we've been writing to their equivalent number representation. And lastly, you'll get to use a little bit of this understanding of instruction representation to decipher your first piece of self modifying code! (optional - but it's awesome so you should totally do it anyways.)

This is also a great opportunity for you to practice commenting your MIPS! Yay documentation! No, but seriously, you should have comments everywhere. It makes it oh so much easier to understand what's going on for the readers, which makes it oh so much more likely for you to get credit for your work! =)

Submit your solution by creating a directory named hw4 that contains files named hw4q1.s, hw4q2.s, hw4q3.s, hw4q4.s, and hw4q5.txt. (Note that capitalization matters in file names; the submission program will not accept your submission if your file names differ at all from those specified.) From within that directory, type "submit hw4". This is not a partnership assignment; hand in your own work.

## Problem 0

Copy the contents of ~cs61c/hw/04 to a suitable location in your home directory.

 \$ cp -r ~cs61c/hw/04/ ~/hw4

## Problem 1

P&H Exercise 2.18.2, both parts (a) and (b). Save your answers in hw4q1.s

## Problem 2

P&H Exercise 2.19.1 and 2.19.2, do part (a) only for both problems. Save your answers in hw4q2.s

## Problem 3

Write the following VectorMult function as a MIPS procedure call:

 ```struct vector { int x; int y; }; void VectorMult(struct vector** vectors, int len, int scale) { ... } ```

The function VectorMult takes an array of pointers to struct vector and multiplies each of the struct vector by scale. The product of vector (x1,y1) and integer scale is simply (x1 * scale, y1 * scale). The function should replace each of the vector in vectors by the corresponding product.

You can assume the integer x is stored at lower memory address than y. Therefore, if the address of a struct vector is at 0x50000, than the member x is located at 0x50000, and the member y is located at 0x50004.

## Problem 4

Write two recursive MIPS functions: one called remove and one called print, that will remove a value and print the contents of, respectively, the linked list set up in hw4q4.s. The skeleton code follows:

 ```.data L9: .word 9 0 L8: .word 8 L9 L7: .word 7 L8 L6: .word 6 L7 L5: .word 5 L6 L4: .word 4 L5 L3: .word 3 L4 L2: .word 2 L3 L1: .word 1 L2 L0: .word 0 L1 .text main: la \$a0 L0 addi \$a1 \$0 9 move \$s0 \$a0 jal remove move \$a0 \$v0 jal print addiu \$v0 \$zero 10 syscall remove: print: ```

(also contained in hw4q4.s)

Especially for remove, you'd probably do yourself a big favor by writing a bit of C code first. A correctly working program will print "012345678" (Note the lack of punctuation or newlines). No, changing around the structure of the defined linked list, while somewhat clever, will not get you credit for this problem - we want to see functioning MIPS code! (and don't hard code the solution to just work for removing a node containing 9!)

## Problem 5

Translate the following MIPS instructions (which copies a string from \$s0 to \$s1) to their corresponing number represtations:

 ```1 main: addiu \$t0 \$zero 0 2 loop: addu \$t1 \$s1 \$t0 3 addu \$t2 \$s2 \$t0 4 lw \$t3 0(\$t1) 5 sw \$t3 0(\$t2) 6 addiu \$t0 \$t0 4 7 bne \$t3 \$zero loop ```

Also included is hw4q5.txt. There a section for each line number which is labeled according to the line number and instruction text. In each section, include the binary representation of the instruction, with spaces demarcating the logical divisions in the instruction. Also include a list of what the names of the different parts of the instruction are followed by the decimal value contained in that section. For example, if this was the line of code I was translating:

 ```1 main: add \$t7 \$t8 \$t9 ```

Then this would be my answer for that line:

 ```================================ 1 main: add \$t7 \$t8 \$t9 ================================ 000000 11000 11001 01111 00000 100000 Type: R op: 0 rs: 24 rt: 25 rd: 15 shamt: 0 funct: 32 ```

## Extra For Experts

Self modifying code! Don't worry, this example won't totally blow your mind:

 ```.data array: .word 0 0 0 0 0 0 0 0 0 0 .text main: la \$s0 array loop: addiu \$a0 \$zero 10 beq \$a0 \$zero Exit addiu \$a0 \$a0 -1 jal morph addu \$t0 \$s0 \$v0 sw \$v0 0(\$t0) j loop Exit: addiu \$v0 \$zero 10 syscall morph: sll \$v0 \$a0 2 la \$t0 loop lw \$t1 0(\$t0) addiu \$t1 \$t1 -1 sw \$t1 0(\$t0) jr \$ra ```

(also contained in hw4q6.s)

For this problem explain what lines accomplish the self modification (and how), what the effect of the modification is, and what the overall behavior of the program is. Place your answers in hw4q6.txt.

NOTE: The MARS simulator has a bug in it that does not allow for self modifying code. If we can fix this we will, but for now, try to run through the code mentally.