exam1-f06
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exam1-f06

Course Number: CS 15-213, Spring 2006

College/University: Carnegie Mellon

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Andrew login ID: Full Name: CS 15-213, Fall 2006 Exam 1 Wednesday October 4, 2006 Instructions: Make sure that your exam is not missing any sheets, then write your full name and Andrew login ID on the front. Write your answers in the space provided below the problem. If you make a mess, clearly indicate your final answer. The exam has a maximum score of 56 points. The problems are of varying difficulty. The...

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login Andrew ID: Full Name: CS 15-213, Fall 2006 Exam 1 Wednesday October 4, 2006 Instructions: Make sure that your exam is not missing any sheets, then write your full name and Andrew login ID on the front. Write your answers in the space provided below the problem. If you make a mess, clearly indicate your final answer. The exam has a maximum score of 56 points. The problems are of varying difficulty. The point value of each problem is indicated. Pile up the easy points quickly and then come back to the harder problems. This exam is OPEN BOOK. You may use any books or notes you like. Calculators are allowed, but no other electronic devices. Good luck! 1 (8): 2 (8): 3 (8): 4 (6): 5 (8): 6 (8): 7 (10): TOTAL (56): Page 1 of 10 Problem 1. (8 points): Assume we are running code on an IA32 machine, which has a 32-bit word size and uses two's complement arithmetic for signed integers. Consider the following definitions: int x = foo(); unsigned ux = x; Fill in the empty boxes in the table below. For each of the C expressions in the first column, either: State that it is true of all argument values, or Give an example where it is not true. Puzzle x < 0 x > 0 x > 0 (x>>31) == -1 x < 0 (x*2) < 0 (x+1) > 0 (~x + 2) <= 0 x < 0U ((x ^ x>>31) + 1) > 0 True / Counterexample False (TMin) ((x>>31)+1) == (x>=0) x >= 0 ((!x - 1) & x) == x ((int)(ux >> 31) + ~0) == -1 -(x | (~x + 1)) > 0 Page 2 of 10 Problem 2. (8 points): Consider the following 5-bit floating point representations based on the IEEE floating point format. This format does not have a sign bit it can only represent nonnegative numbers. There are k = 3 exponent bits. The exponent bias is 3. There are n = 2 fraction bits. Numeric values are encoded as a value of the form V = M 2E , where E is exponent after biasing, and M is the significand value. The fraction bits encode the significand value M using either a denormalized (exponent field 0) or a normalized representation (exponent field nonzero). Below, you are given some decimal values, and your task it to encode them in floating point format. If rounding is necessary, you should use round-to-even, as you did in Lab 1 for the float i2f puzzle. In addition, you should give the rounded value of the encoded floating point number. Give these as whole numbers (e.g., 17) or as fractions in reduced form (e.g., 3/4). Value 9/32 7/8 15/16 9 10 Floating Point Bits 001 00 Rounded value 1/4 Page 3 of 10 Problem 3. (8 points): Consider the following C function's x86-64 assembly code: # On entry %edi = n # 0000000004004a8 <foo>: 4004a8: b8 00 00 00 00 4004ad: 83 ff 01 4004b0: 7e 1a 4004b2: 01 f8 4004b4: ba 00 00 00 00 4004b9: 39 fa 4004bb: 7d 08 4004bd: 01 d0 4004bf: ff c2 4004c1: 39 fa 4004c3: 7c f8 4004c5: ff cf 4004c7: 83 ff 01 4004ca: 7f e6 4004cc: f3 c3 Please fill in the corresponding C code: mov $0x0,%eax cmp $0x1,%edi jle 4004cc <foo+0x24> add %edi,%eax mov $0x0,%edx cmp %edi,%edx jge 4004c5 <foo+0x1d> add %edx,%eax inc %edx cmp %edi,%edx jl 4004bd <foo+0x15> dec %edi cmp $0x1,%edi jg 4004b2 <foo+0xa> repz retq # treat repz as a no-op int foo (int n) { int a, i; a = 0; for (; n > _____; _____) { a = a + _____; for (i = _____; i < _____; _____) a = a + _____; } return _____; } Page 4 of 10 Problem 4. (6 points): Consider the C code below, where H and J are constants declared with #define. int array1[H][J]; int array2[J][H]; int copy_array(int x, int y) { array2[y][x] = array1[x][y]; return 1; } Suppose the above C code generates the following x86-64 assembly code: # On entry: # %edi = # %esi = # copy_array: movslq movslq movq salq subq addq leaq addq movl movl movl ret x y %esi,%rsi %edi,%rdi %rsi, %rax $7, %rax %rsi, %rax %rdi, %rax (%rdi,%rdi,2), %rdi %rsi, %rdi array1(,%rdi,4), %edx %edx, array2(,%rax,4) $1, %eax What are the values of H and J? H = J = Page 5 of 10 Problem 5. (8 points): Consider the following C declaration: typedef struct WineNode { int vintages[3]; double cost; char z; WineNode *next; short ages[5]; int type; char a; } WineNode; A. Using the template below (allowing a maximum of 80 bytes), indicate the allocation of data for the struct WineNode. Mark off and label the areas for each element (arrays may be labeled as a single element). hatch Cross the parts that are allocated, but not used. Clearly mark the end of the struct. Assume the 64 bit alignment rules and X86-64 data structure sizes discussed in class. WineNode: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ Page 6 of 10 B. How many bytes of space in WineNode are wasted?______________ C. Now rewrite the WineNode struct in the space provided below so that the amount of wasted allocated space in WineNode is minimized. typedef struct WineNode { } WineNode; D. Now rewrite the allocation for WineNode as you did before using this new specification. WineNode: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ E. How many bytes of space in this new WineNode are wasted?______________ Page 7 of 10 Problem 6. (8 points): Consider the following data structure declarations: struct node { struct data d; struct node *next; }; struct data { int x; char str[6]; }; Below are given four C functions and four x86-64 code blocks. Next to each of the x86-64 code blocks, write the name of the C function that it implements. int alpha(struct node *ptr) { return ptr->d.x; } movq addq ret 16(%rdi), %rax $4, %rax char *beta(struct node *ptr) { ptr = ptr->next; return ptr->d.str; } movq ret %rdi, %rax char gamma (struct node *ptr) { return ptr->d.str[4]; } movl ret (%rdi), %eax int *delta (struct node *ptr) { struct data *dp = (struct data *) ptr; return &dp->x; } movsbl ret 8(%rdi),%eax Page 8 of 10 Reverse Engineering Switch Code The next problem concerns the code generated by GCC for a function involving a switch statement. Following a bounds check, the code uses a jump to index into the jump table 400476: ff 24 d5 a0 05 40 00 jmpq *0x4005a0(,%rdx,8) Using GDB, we extract the 8-entry jump table as: 0x4005a0: 0x4005b0: 0x4005c0: 0x4005d0: 0x0000000000400480 0x0000000000400480 0x0000000000400480 0x0000000000400485 0x0000000000400491 0x0000000000400496 0x0000000000400489 0x0000000000400496 The following block of disassembled code implements the branches of the switch statement 400480: 400484: 400485: 400489: 40048c: 40048f: 400490: 400491: 400495: 400496: 40049a: 48 c3 48 48 48 90 c3 48 c3 48 c3 8d 04 3f 0f af f7 89 f8 21 f0 lea retq imul mov and nop retq lea retq lea retq (%rdi,%rdi,1),%rax %rdi,%rsi %rdi,%rax %rsi,%rax 8d 04 37 8d 46 ff (%rdi,%rsi,1),%rax 0xffffffffffffffff(%rsi),%rax Page 9 of 10 Problem 7. (10 points): Fill in the blank portions of the C code below to reproduce the function corresponding to this object code. You can assume that the first entry in the jump table is for the case when s equals 0. Parameters a, b, and s are passed in registers %rdi, %rsi, and %rdx, respectively. long fun(long a, long b, long s) { long result = 0; switch (s) { case ___: case ___: result = _____; break; case ___: b = _____; /* Fall through */ case ___: result = _____; break; case ___: result = _____; break; default: result = _____; } return result; } Page 10 of 10

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