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

Transcription

1 CPSC 213, Summer 2017, Term 2 Midterm Exam Solution Date: July 27, 2017; Instructor: Anthony Estey 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. 1000: 0x : 0x : 0x : 0x : 0x : 0x : 0x : 0x : 0x12 100c: 0x00 100d: 0x00 100e: 0x10 100f: 0x00 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). Yes. 1004: 0x : 0x : 0x : 0x11 100c: 0x00 (the values for each of the 4 bytes in memory 100d: 0x10 would be in opposite order, but technically 100e: 0x00 only the values 100d and 100e actually change 100f: 0x00 in this case)

2 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: ld $0x2000, r1 ld $5, r2 ld $4, r3 add r2, r3 st r3, 0(r1) Memory address written to: 0x2000 Value written to memory: 9 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) 0x = 0x216 in hexadecimal = 534 in decimal 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]? 8, 4, 3, 2, 1 3b What is the value of a? 0x3000 3c What is the value of *a? 8 3d What is the value of *a + *b + *c? 18 2

3 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. ld $i, r0 # r0 = &i ld (r0), r0 # r0 = i ld $a, r1 # r1 = &a st r0, (r1, r0, 4) # a[i] = i 4b Give assembly code for the C statement: i = a[i+b[i]]. ld $i, r0 # r0 = &i ld (r0), r1 # r1 = i ld $b, r2 # r2 = &b; ld (r2), r2 # r2 = b[0] ld (r2, r1, 4), r2 # r2 = b[i] add r1, r2 # r2 = i + b[i] ld $a, r3 # r3 = &a ld (r3, r2, 4), r3 # r3 = a[i+b[i]] st r3, (r0) # 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 Yes ld $i, r0 # r0 = &i ld (r0), r0 # r0 = i ld $j, r1 # r1 = &j ld (r1), r1 # r1 = j add r1, r0 # r0 = i + j ld $k, r1 # r1 = $k ld (r1), r1 # r1 = k add r1, r0 # r0 = i + j + k ld $a, r1 # r1 = &a ld (r1, r0, 4), r1 # r1 = a[i + j + k] ld $i, r0 # r0 = &i st r1, (r0) # i = a[i + j + k] Solution using more than 2 regs: ld $a, r0 # r0 = &a ld $i, r1 # r1 = &i ld (r1), r2 # r2 = i ld $j, r3 # r3 =&j ld (r2), r3 # r3 = j ld $k, r4 # r4 = &k ld (r3) r4 # r4 = k add r3, r2 # r2 = i + j add r4, r2 # r2 = i + j + k ld (r0, r2, 4), r5 # r5 = a[i + j + k] st r5, (r1) # i = a[i + j + k] 3

4 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? 12 Comments are not 5b Give assembly code for the C statement: other.id = root->node.id;. ld $other, r0 # r0 = &other ld $root, r1 # r1 = &root ld (r1), r1 # r1 = root ld 20(r1), r1 # r1 = root->node.id st r1, 12(r0) # other.id = root->node.id 5c Give assembly code for the C statement: root->right->node.id = root->left->right->node.parent->node.id;. ld $root, r0 # r0 = &root ld (r0), r0 # r0 = root mov r0, r1 # r1 = root ld 4(r0), r0 # r0 = root->right ld (r1), r1 # r1 = root->left ld 4(r1), r1 # r1 = root->left->right ld 16(r1), r1 # r1 = root->left->right->parent ld 20(r1), r1 # r1 = root->left->right->parent->node.id st r1, 20(r0) # root->r->n.id = root->l->r->p->n.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; Reads (3): root root->left root->left->node.id. 4

5 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; ld $x, r0 # r0 = &x ld 0x0(r0), r1 # r1 = x ld $y, r2 # r2 = &y ld 0x0(r2), r3 # r3 = y mov r3, r4 # r4 = y not r4 inc r4 # r4 = -y mov r4, r5 # r5 = -y add r1, r4 # r4 = x-y beq r4, L0 # goto L0 if x == y st r3, 0x0(r0) # x = y br L1 L0: st r5, 0x0(r0) # x = -y L1: halt 5

6 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: Circle the C snippet your assembly code translates: 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 ld (r0), r1 # r1 = x = x ld $y, r2 # r2 = &y ld (r2), r3 # r3 = y = y L0: beq r1, L1 # goto L1 if x == 0 mov r3, r4 # r4 = y add r4, r3 # r3 = y + y dec r3 dec r1 br L0 # goto L0 L1: st r3, (r2) # y = y st r1, (r0) # x = x 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--; 6

7 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; Memory leak; malloc d item is not freed and pointer is lost when assigned to new malloc d item. 7b void foo() { a = NULL; *a = 2; Code will crash; trying to dereference a after pointing it to NULL. 7c void foo() { free(a); int *b = a;... Dangling pointer; b is assigned to a memory location that was just freed. 7d void foo() { free(a); a = NULL; No bug. 7

8 7e void bar(int *b) {... void foo() { bar(a); free(a);... Possible dangling pointer; bar might save a pointer that is then freed in foo 7f void bar(int *b) { int *c = b; void foo() { bar(a); free(a); No bug. 7g void bar(int *b) { b = malloc(sizeof(int)); void foo() { void foo() { bar(a); free(a); a = NULL; Memory leak; there is a memory allocation in bar assigned to a local variable that is lost when bar finishes. 8

9 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). 4 8b Determine what the reference count should be output by the printf statement if reference counting had been correctly implemented. 1 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) { if (adder) dec_ref(adder); 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); //inc_ref(j); return i; int main (void) { struct item *i1 = make_item(); //inc_ref(i1); struct item *i2 = make_item(); //inc_ref(i2); printf("i1 s ref count: %d", i1->ref_count); //What should i1->ref_count be here?... 9

10 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 # r2 = &a ld (r2), r0 # r0 = a ld 4(r2), r1 # r1 = b ld $0, r5 # r5 = temp_c mov r1, r3 # r3 = b not r3 inc r3 # r3 = -b L1: mov r0, r4 # temp_a = temp_a add r3, r4 # temp_a = temp_a - b bgt r4, L2 # goto L2 if temp_a > b beq r4, L2 # goto L2 if temp_a >= b else: br L3 # else goto L3 L2: inc r5 # temp_c++ add r3, r0 # temp_a -= b br L1 # goto L1 L3: st r0, (r2) # a = temp_a ld $c, r2 # r2 = &c st r5, (r2) # c = temp_c 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): c = 0; int i = 0; while (a >= b) { while (a >= b) { c++; i++; a -= b; a-=b; c = i; 9b What are the final values at addresses 0x2000, 0x2004, and 0x2008? 2, 6, and 3. 9c Write the code as a single mathematical expression using a, b, and c. Optionally: c = a/b a = a%b 9d Explain what the code does in one sentence (a purpose statement). Dividing a by b, storing remainder in a, and result in c (rounded down to the nearest integer). 10