CPSC 213, Summer 2017, Term 2 Midterm Exam Date: July 27, 2017; Instructor: Anthony Estey

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1 CPSC 213, Summer 2017, Term 2 Midterm Exam Date: July 27, 2017; Instructor: Anthony Estey 1. This is a closed book exam. No notes. No electronic calculators. 2. Answer in the space provided. Show your work; use the backs of pages if needed 3. You may detach the last page of the exam. It is an ISA reference sheet. 4. You have 2 hours to write the exam, good luck! Answer in the space provided. Show your work; use the backs of pages if needed. There are 9 questions on 10 pages, totaling 85 marks. You have 2 hours to complete the exam. STUDENT NUMBER: NAME: Q1 / 10 Q2 / 6 Q3 / 8 Q4 / 8 Q5 / 10 Q6 / 8 Q7 / 14 Q8 / 6 Q9 / 15 SIGNATURE:

2 1 (10 marks) Memory and Numbers. Consider the following C code containing global variables a, b, c, and d, that is executed on a big endian, 32-bit processor. Assume that the address of a is 0x1000 and that the compiler allocates the variables contiguously, in the order they appear, wasting no memory between them other than what is required to ensure that they are properly aligned (and assuming that int s and pointers are 4-bytes long). With this information, you can determine the value of certain bytes of memory following the execution of foo(). int a; int b; char c; int* d; void foo() { a = 0x ; c = 0x12; d = &a; b = *d; *d = 0x ; 1a List the address and value of every memory location whose address and value you know. Use the form address: value. List every byte on a separate line and list all numbers in hex. 1b Would anything change about your answer to part A if the processor were little-endian (yes/no). If yes, list the address: value pairs for only memory addresses that would be changed (if any). 2

3 2 (6 marks) Machine Model and Instructions In assignment 2, you had to complete the CPU.java program using the tables (found on the last page of this exam) as a reference for the SM213 instruction semantics. 2a First, translate the Machine Language instructions below into assembly. Use the tables to determine what value is written to memory, and what address it is written to after the instructions have been executed. Machine Language Instructions: Assembly Language Instructions: Memory address written to: Value written to memory: 2b Assume the address of the first instruction is at memory address 0x200 (0x200 in base-16 translates to 512 in base-10/decimal). What is the address of the first instruction after the final instruction in the snippet above? Listing the sizes of each instruction may help you come up with the answer, and could award you part marks for an incorrect solution. (You can answer in decimal or hexadecimal) 3

4 3 (8 marks) C Pointers. Consider the following global variable declarations int *a; int *b; int c[5] = (int[5]){5, 4, 3, 2, 1; Assume that the address of a is at location 0x1000, the address of b is at location 0x2000, and the address of c is at location 0x3000. Now, consider the the execution of the following additional code: a = &c[1]; b = c + 3; *c = *a + *b; a = a - 1; *c = *a + *b; Answer these questions about the value of these variables following the execution of the code. 3a What are the values in array c, c[0]... c[4]? 3b What is the value of a? 3c What is the value of *a? 3d What is the value of *a + *b + *c? 4

5 4 (8 marks) Global Variables and Arrays. Answer the following questions about the global variables a, b, i, j, and k, declared as follows. Treat each sub-question separately (i.e., do not use values computed in prior sub-questions). Comments are not required, but they will probably help you. int i, j, k; int a[10]; int* b; 4a Give assembly code for the C statement: a[i] = i. 4b Give assembly code for the C statement: i = a[i+b[i]]. 4c Give assembly code for the C statement: i = a[i + j + k] (use as many registers as you want). Could the C statement be written if the system only had 2 registers available for use? YES / NO 5

6 5 (10 marks) Structs and Instance Variables. Assume two global variables root and other, have been declared. struct B { int data[2]; struct A* parent; int id; ; struct A { struct A* left; struct A* right; struct B node; ; struct A* root; struct B other; Treat each sub-question separately (i.e., do not use values computed in prior sub-questions). required, but they will probably help you. Assume all pointers are 4 bytes long. 5a What is the offset to id in struct B? Comments are not 5b Give assembly code for the C statement: other.id = root->node.id;. 5c Give assembly code for the C statement: root->right->node.id = root->left->right->node.parent->node.id;. 5d Determine how many memory reads are required to execute the C statement: (Listing the read operations in Assembly or C is not required by may help you, and get you part marks). other.id = root->left->node.id; 6

7 6 (8 marks) Static Control Flow. Answer these questions where x and y are global integer variables, and have been initialized to values greater than 0. 6a Give assembly code for the following C code snippet: if (x == y) x = -y else x = y; 7

8 6b Imaginary Scenario: Your partner sent you their part of the assignment, but somehow all of the lines of assembly code got scrambled (darn vim!). The code you received has all of the necessary assembly instructions, but they are out of order. Additionally, your partner could not remember what the code was supposed to do. Your first task is to reorganize the given assembly instructions on the left to produce a correct translation of one of the four code snippets on the right. Each line of scrambled code can only be used once. Once complete, circle the snippet on the right that the assembly code correctly translates. (Hint: Think systematically. The code snippets on the right are all similar, so think about how you would begin to translate any one of them. Cross off one line from the supplied code when you use it in your solution. I have started the process for you, and crossed off ld $x, r0) Comments are not required, but they will probably help you. Code supplied by partner: mov r3, r4 L1: st r3, (r2) ld (r0), r1 br L0 ld (r2), r3 dec r3 L0: beq r1, L1 ld $x, r0 st r1, (r0) add r4, r3 ld $y, r2 dec r1 Write your (unscrambled) code here: ld $x, r0 # r0 = &x Circle the C snippet your assembly code translates: while (x > 0) { x += y; x--; y--; while (x > 0) { y += y; x--; y--; while (x > 0) { x += x; x--; y--; while (x > 0) { y += x; x--; y--; 8

9 7 (14 marks) Dynamic Allocation. Consider each of the following pieces of C code to determine whether it contains (or may contain) a memory leak, dangling pointer, or other bug that could crash the program. If you think the code snippet does contain a bug, clearly state which bug it is (or may be), and explain why the code produces the bug. If you are unable to determine for certain that the bug exists, but have indicated that it might, clearly explain this as well. Do not fix the bug. If no bug exists, simply say so. Most of the code snippets are very similar. I have highlighted changes from previous versions and/or key things to look for in bold font. 7a void foo() { a = malloc(sizeof(int)); *a = 2; 7b void foo() { a = NULL; *a = 2; 7c void foo() { free(a); int *b = a;... 7d void foo() { free(a); a = NULL; 9

10 7e void bar(int *b) {... void foo() { bar(a); free(a);... 7f void bar(int *b) { int *c = b; void foo() { bar(a); free(a); 7g void bar(int *b) { b = malloc(sizeof(int)); void foo() { void foo() { bar(a); free(a); a = NULL; 10

11 8 (6 marks) Reference Counting. The following code attempts to implement reference counting, but the reference counts are not handled correctly. 8a What is the reference count value output (in the bold printf statement). 8b Determine what the reference count should be output by the printf statement if reference counting had been correctly implemented. 8c Modify the code (mostly by adding/removing add_ref and/or dec_ref procedure calls), so the program correctly implements reference counting. struct item { int ref_count; int data; ; struct item *adder = NULL; int number = 0; void inc_ref(struct item *i) { i->ref_count++; void dec_ref(struct item* i) { i->ref_count--; if (i->ref_count == 0) free(i); struct item *process(struct item *temp) { adder = temp; inc_ref(adder); adder->data = adder->data + 1; return adder; struct item *make_item() { struct item* i = (struct item*)malloc(sizeof(struct item)); i->ref_count = 1; i->data = number++; struct item *j = process(i); return i; int main (void) { struct item *i1 = make_item(); struct item *i2 = make_item(); printf("i1 s ref count: %d", i1->ref_count); //What should i1->ref_count be here?... 11

12 9 (15 marks) Reading Assembly. Comment the following assembly code and then translate it into C. Use the back of the preceding page for extra space if you need it..pos 0x100 ld $a, r2 # ld (r2), r0 # ld 4(r2), r1 # ld $0, r5 # mov r1, r3 # not r3 inc r3 # L1: mov r0, r4 # add r3, r4 # bgt r4, L2 # beq r4, L2 # else: br L3 # L2: inc r5 # add r3, r0 # br L1 # L3: st r0, (r2) # ld $c, r2 # st r5, (r2) # halt.pos 0x1000 a:.long 20 # assume a >= 0 b:.long 6 # assume b > 0 c:.long 20 9a Translate into C (you do not need to include variable declarations): 9b What are the final values at addresses 0x2000, 0x2004, and 0x2008? 9c Write the code as a single mathematical expression using a, b, and c. 9d Explain what the code does in one sentence (a purpose statement). 12

13 You may remove this page. These two tables describe the SM213 ISA. The first gives a template for instruction machine and assembly language and describes instruction semantics. It uses s and d to refer to source and destination register numbers and p and i to refer to compressed-offset and index values. Each character of the machine template corresponds to a 4-bit, hexit. Offsets in assembly use o while machine code stores this as p such that o is either 2 or 4 times p as indicated in the semantics column. The second table gives an example of each instruction. Operation Machine Language Semantics / RTL Assembly load immediate 0d-- vvvvvvvv r[d] vvvvvvvv ld $vvvvvvvv,rd load base+offset 1psd r[d] m[(o = p 4) + r[s]] ld o(rs),rd load indexed 2bid r[d] m[r[b] + r[i] 4] ld (rb,ri,4),rd store base+offset 3spd m[(o = p 4) + r[d]] r[s] st rs,o(rd) store indexed 4sdi m[r[b] + r[i] 4] r[s] st rs,(rb,ri,4) halt F000 (stop execution) halt nop FF00 (do nothing) nop rr move 60sd r[d] r[s] mov rs, rd add 61sd r[d] r[d] + r[s] add rs, rd and 62sd r[d] r[d] & r[s] and rs, rd inc 63-d r[d] r[d] + 1 inc rd inc addr 64-d r[d] r[d] + 4 inca rd dec 65-d r[d] r[d] 1 dec rd dec addr 66-d r[d] r[d] 4 deca rd not 67-d r[d]!r[d] not rd shift 7dss r[d] r[d] << ss shl ss, rd (if ss is negative) shr -ss, rd branch 8-pp pc (aaaaaaaa = pc + pp 2) br aaaaaaaa branch if equal 9rpp if r[r] == 0 : pc (aaaaaaaa = pc + pp 2) beq rr, aaaaaaaa branch if greater Arpp if r[r] > 0 : pc (aaaaaaaa = pc + pp 2) bgt rr, aaaaaaaa jump B--- aaaaaaaa pc aaaaaaaa j aaaaaaaa get program counter 6Fpd r[d] pc + (o = 2 p) gpc $o, rd jump indirect Cdpp pc r[d] + (o = 2 pp) j o(rd) jump double ind, b+off Cdpp pc m[(o = 4 pp) + r[d]] j *o(rd) jump double ind, index Edi- pc m[4 r[i] + r[d]] j *(rd,ri,4) Operation Machine Language Example Assembly Language Example load immediate ld $0x1000,r1 load base+offset 1123 ld 4(r2),r3 load indexed 2123 ld (r1,r2,4),r3 store base+offset 3123 st r1,8(r3) store indexed 4123 st r1,(r2,r3,4) halt f000 halt nop ff00 nop rr move 6012 mov r1, r2 add 6112 add r1, r2 and 6212 and r1, r2 inc 6301 inc r1 inc addr 6401 inca r1 dec 6501 dec r1 dec addr 6601 deca r1 not 6701 not r1 shift 7102 shl $2, r1 71fe shr $2, r1 branch 1000: 8003 br 0x1008 branch if equal 1000: 9103 beq r1, 0x1008 branch if greater 1000: a103 bgt r1, 0x1008 jump b j 0x1000 get program counter 6f31 gpc $6, r1 jump indirect c104 j 8(r1) jump double ind, b+off d102 j *8(r1) jump double ind, index e120 j *(r1,r2,4) 13