Final Review 09

7/23/2019 Final Review 09

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Question 1

int decode2(int x, int y, int z)

1: movl 16(%ebp), %eax2: movl 12(%ebp), %edx3: subl %eax, %edx4: movl %edx, %eax5: imull 8(%ebp), %edx6: sall $31, %eax7: sarl $31, %eax8: xorl %edx, %eax

Q. Write the corresponding C code

Answer:

int decode2(int x, int y, int z) {int t1 ! " #

int t2 x t1int t3 (t1 && 31) '' 31int t4 t3 t2return t4

}


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Question 21: int absdi2(int x, int !) *2: int result3: i (x & !)4: result ! + x5: else6: result x + !7: return result8:

Corresponding assemby code

1: movl 8(%ebp), %edx2: movl 12(%ebp), %e-x3: movl %edx, %eax

4: subl %e-x, %eax5: -mpl %e-x, %eax6: ./e 037: movl %e-x, %eax8: subl %edx, %eax: 03

a. What s!btractions are per"ormed when x # y$ When x%& y$

b. 'n what way does this code deiate "rom the standard impementation o" an i"ese$

c. *sing C syntax (inc!ding goto+s), show the genera "orm o" this transation

d. What restrictions mustbe imposed on the !se o this transation to g!arantee that it has

the behaior speci"ied by the C code$

Answers:

a. when x # y it "irst comp!tes x y and then y x, when x %& y it comp!tes y x.

b. -he code "or the thenstatement is exec!ted !nconditionay.

c.

thenstatement

t & testexpression

i" (t)

goto done

esestatement


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done:

Question 3:

witch statements are partic!ary tric/y to reerse engineer "rom the ob0ect code. 'n the

"oowing proced!re, the body o" the switch statement has been remoed:

1: int sit-prob(int x) *2: int result x3: sit- (x) *4: ill in -ode ere 5: 6: return result7:

1ig!re .3 shows the disassembed ob0ect code "or the proced!re. We are ony

interested in the part o" the code shown on ines 4 thro!gh 56. We can see on ine 4 that

parameter x (at o""set 3 reatie to 7ebp) is oaded into register 7eax, corresponding toprogram ariabe res!t. -he 8ea 9x9(7esi), 7esi instr!ction on ine 55 is a nop

instr!ction inserted to ma/e the instr!ction on ine 52 start on an address that is a

m!tipe o" 56.

-he 0!mp tabe resides in a di""erent area o" memory. *sing the deb!gger ;6w0x8048468: 0x080483d5 0x080483eb 0x080483d5 0x0x80483e0

0x8048478: 0x080483e5 0x080483e8

(gdb)

Assemby Code:1: 989483-9 &sit-prob':2: 89483-9: pus %ebp3: 89483-1: mov %esp,%ebp4: 89483-3: mov 9x8(%ebp),%eax5: 89483-6: lea 9x-e(%eax),%edx6: 89483-: -mp $9x5, %edx7: 89483--: .a 89483eb &sit-prob9x2b'8: 89483-e: .mp 9x8948468(,%edx,4)

: 89483d5: sl $9x2, %eax19: 89483d8: .mp 89483ee &sit-prob9x2e'11: 89483da: lea 9x9(%esi),%esi12: 89483e9: sar $9x2, %eax13: 89483e3: .mp 89483ee &sit-prob9x2e'14: 89483e5: lea (%eax, %eax, 2), %eax15: 89483e8: imul %eax, %eax16: 89483eb: add $9xa, %eax17: 89483ee: mov %ebp, %esp18: 894839: pop %ebp1: 894831: ret


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29: 894832: mov %esi, %esi

;nser:int sit-prob(int x) * int result x

sit- (x) *-ase 59:-ase 52: result && 2brea


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Question 4:t!pede stru-t *

int letastru-t a=>?@Aint ri/t

bstru-t

void test(int i, bstru-t bp) * astru-t ap Bbp"'a=iA int n bp"'let bp"'ri/t ap"'x=ap"'idxA n

*n"ort!natey, the 8.h "ie de"ining the compietime constant C?- and the str!ct!re

a@str!ct are in "ies "or which yo! do not hae access priieges. 1ort!natey yo! hae

access to the a 8.o ersion o" the code, which yo! are abe to disassembe with theob0d!mp program, yieding the disassemby shown beow.

9 99999999&test'

5 9: pus %ebp

2 1: mov %esp, %ebp

3: pus %ebx

4 4: mov 9x8(%ebp), %eax

7: mov 9x-(%ebp), %e-x

6 a: lea (%eax, %eax, 4), %eax

B d: lea 9x4(%e-x, %eax, 4), %eax3 11: mov (%eax), %edx

13: sl $9x2, %edx

59 16: mov 9xb8(%e-x), %ebx

55 1-: add (%e-x), %ebx

52 1e: mov %ebx, 9x4(%edx, %eax, 1)

5 22: pop %ebx

54 23: mov %ebp, %esp

5 25: pop %ebp

56 26: ret


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Answers:

a. C?- is

b.t!pede stru-t * int idx

int x=4A astru-t


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Question 5:union ele *

stru-t *int pint !

e1stru-t *

int xunion ele next

e2

-his decaration i!strates that str!ct!res can be embedded within !nions.

-he "oowing proced!re (with some expressions omitted) operates on a in/ed ist

consisting o" these !nions as ist eements:

void pro- (union ele up)*

up C (up C ) " up C

a. What wo!d be the o""sets (in bytes) o" the "oowing "ieds:

e5.p:

e5.y:

e2.x:

e2.next:

b. Dow many tota bytes wo!d the str!ct!re reE!ire$

c. -he compie generates the "oowing assemby code "or proc:

1 movl 8(%ebp), %eax2 movl 4(%eax), %edx3 movl (%edx), %e-x

4 movl %ebp, %esp5 movl 8(%eax), %eax6 movl 8(%e-x), %e-x7 subl %eax, %e-x8 movl %e-x, 4(%edx)

*se this code to "i in the ban/s "or the C so!rce. -here is ony one answer that does

not per"orm any casting and does not ioate any type constraints.


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Answers:

-his probem ery ceary shows that !nions are simpy a way to associate m!tipe

names (and types) with a singe storage ocation.

a.


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Question 6:

-his E!estion is testing yo!r !nderstanding o" the stac/ "rame str!ct!re.

a. -he "oowing memory image embeds a binary tree with root node at 0x804961c.

Hease draw the ogica organization o" the tree in the same "ormat as the shown

exampe. Hease indicate the address and /ey a!e (in hexidecima) o" a the tree nodesand the pointers "rom parent nodes to chid nodes. -he decaration o" the tree node

str!ct!re is as "oows.

struct tree_node {

int key;

struct tree_node *left;

struct tree_Node *right;

};

/* address of the root */

tree_node *root;

Memory:

0x80495f8: 0x0000000c

0x80495fc: 0x00000000

0x8049600: 0x00000000

0x8049604: 0x0000001f

0x8049608: 0x080495f8

0x804960c: 0x08049610

0x8049610: 0x00000022

0x8049614: 0x00000000

0x8049618: 0x000000000x804961c: 0x00000037

0x8049620: 0x08049604

0x8049624: 0x08049628

0x8049628: 0x0000003c

0x804962c: 0x00000000

0x8049630: 0x08049634

0x8049634: 0x0000004e

0x8049638: 0x00000000

0x804963c: 0x00000000

1: struct tree_node * search(struct tree_node * node, int value)

2: {3: if (node->key == value)

4: return node;

5: else if (node->key > value) {

6: if (node->left == NULL)

7: return NULL;

8: else

9: return search(node->left, value);

10: } else {

11: if (node->right == NULL)

12: return NULL;


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13: else

14: return search(node->right, value);

15: }

16: }

b. !ppose we ca search(root, 9x4e). 1i in the ban/s the a!e o" these memory

ocation so that it shows the stac/ when the exec!tion is at ine 4. (Iore space than

needed is proided. ) Jo! can ass!me that the stac/ stores ony arg!ments, ret!rnaddress, and the ebp register a!e. -he a!e o" ebp is 9xb""""339 when the program

cas the "!nction. Write rtn addr "or ret!rn addresses.

Address Ka!e9xb899 9x4e9xb7- 9x89461-9xb78 rtnaddr9xb74

9xb799xb7e-9xb7e89xb7e49xb7e99xb7d-9xb7d89xb7d49xb7d99xb7--

9xb7-89xb7-4

Answers:

a. (0x804961c/0x37)

/ \

(0x8049604/0x1f) (0x8049628/0x3c)

/ \ \

(0x80495f8/0xc) (0x8049610/0x22) (0x8049634/0x4e)

b.

0xbffff800 0x4e

0xbffff7fc 0x804961c

0xbffff7f8 rtn_addr

0xbffff7f4 0xbffff880

0xbffff7f0 0x4e


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0xbffff7ec 0x8049628

0xbffff7e8 rtn_addr

0xbffff7e4 0xbffff7f4

0xbffff7e0 0x4e

0xbffff7dc 0x8049634

0xbffff7d8 rtn_addr

0xbffff7d4 0xbffff7e4

Question 7:

a.

Which o" the "oowing x36 instr!ctions can be !sed to add two registers and store the

res!t witho!t oerwriting either o" the origina a!es$

(a) mo

(b) add

(c) ea

(d) ?one o" aboe

b.

-he register rax is c!rrenty storing a ?*FF pointer. Which o" the "oowing x36

instr!ctions wi ca!se a segmentation "a!t beca!se o" an inaid memory access$

(a) mo (7rax), 7rcx

(b) ea (7rax), 7rcx

(c) ?one o" the aboe

c. A programmer wishes to compare the contents o" a string caed my@str to the string8;L-. he writes the "oowing C code:

i (m!str DEF@D) 000

Which o" the "oowing appy$

(a) my str is a pointer to the "irst character o" a string in memory

(b) my str is the AC'' a!e o" the "irst character o" a string in memory

(c) my str is a register containing a o" the characters in the string

(d) 8;L- wi compie to a pointer to a string in memory

(e) 8;L- wi compie to the AC'' a!e "or the etter ;

(") 8;L- wi compie to a register containing the string 8;L- represented as an

integer

(g) -he comparison wi aways wor/ as expected

(h) -he comparison wi not necessariy wor/ as expected

(i) -he comparison itse" wi ca!se the program to crash


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d. -he "!nction foo() is decared in a C program as "oows: void foo(int

int_param, char *str_param);

A programmer cas foo() "rom within the "!nction bar() as "oows:

foo(my_int, my_string);

Which o" the "oowing is>are tr!e:

(a) '" foo() changes the a!e o" int param, the change wi propagate bac/ to the

caing "!nctionbar(), in other words, the a!e o" my int wi aso change.

(b) '" foo() changes the second character o" str param, the change wi propagate

bac/ to the caing "!nction bar(), in other words, the second character o" my

string wi aso change.

(c) '" foo() changes the address o" str param to point to a di""erent string, the

change wi propagate bac/ to the caing "!nction bar(), in other words, my string

wi now point to a di""erent string.

(d) ?one o" the aboe.

e.A programmer has decared an array in a C program as "oows:

int my_array[100];

Which o" the "oowing gie(s) the address o" the eighth eement in the array (bearing in

mind that the "irst eement in the array is at index zero):


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(a) my array[7]

(b) &my array[7]

(c) my array + 7

(d) my array + 28

(e) ?one o" the aboe

g. A programmer has stored an 3bit a!e in memory. -he pointer:

char GptrM

points to the ocation where it is stored. De or she now wants to retriee the a!e and

store it into the ariabe:

int a!eM

Which o" the "oowing (i" any) wi achiee this propery$

(a) a!e & ptrM

(b) a!e & GptrM

(c) a!e & (int)ptrM

(d) a!e & (int G)ptrM

(e) a!e & G(int G)ptrM

(") ?one o" the aboe

h. ;ien code "or an impicit ist memory aocator, yo! caim yo! can improe

per"ormance by creating a in/ed ist o" "ree boc/s. Why does this hep increase

per"ormance$

(a) -raersing a in/ed ist is signi"icanty "aster than moing "rom boc/ to boc/ in the

impicit ist.

(b) -he impicit ist had to inc!de eery boc/ in memory, b!t the in/ed ist co!d 0!st

inc!de the "ree


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boc/s.

(c) -he compier /nows how to optimize the code "or a in/ed ist by !nroing oops,

b!t wasnNt abe to do

this "or the impicit ist.

(d) Daing a in/ed ist made coaescing signi"icanty "aster.

(e) ?one o" the aboe.

Answer(s):

Answers:


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Question 8: Synhroni!ation

A m!tithreaded program has two goba data str!ct!res that wi be shared among the

threads. -he data str!ct!res are not necessariy accessed at the same time. Which o" the

"oowing is>are tr!e (i" any)$

(a) '" the program has ony one semaphore, and threads ca P on that singe semaphore

be"ore !sing either

o" the data str!ct!res, the code wi not wor/ correcty.

(b) Daing one semaphore wi wor/, b!t haing two, one per shared data str!ct!re, may

aow "or increased

per"ormance.

(c) '" the machine has ony one processor, ony one o" the threads can r!n at a time, so

semaphores are notnecessary in that case.

(d) ?one o" the aboe.

"orret answer(s):

Answers:


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Question #: Synhroni!ation

A barbershop consists o" n waiting chairs and the one barber chair. '" there are no

c!stomers, the barber waits. '" a c!stomer enters, and a the waiting chairs are occ!pied,

then the c!stomer eaes the shop. '" the barber is b!sy, b!t waiting chairs are aaiabe,

then the c!stomer sits in one o" the "ree waiting chairs. We proide yo! the s/eeton

code witho!t any synchronization.extern int ?

int -ustomers 9

void -ustomer() *

i (-ustomers ' ?) *

return ?G

-ustomers 1

/etHair>ut()

-ustomers " 1

return ?G

void barber() *ile (1) *

-utHair()

O!r so!tion !ses threebinary sempahores:mutex: to ontro$ aess to the g$oba$ %ariab$e ustomers

-ustomer: to signa$ a ustomer is in the sho&

barber: to signa$ the barber is busy

a. What are the initia a!es "or the three semaphores$

b. Compete the aboe witho!t changing A?J existing code b!t ony by "iing in as

many copies o" the "oowing ines as yo! need.

I(Bmutex)J(Bmutex)I(B-ustomer)J(B-ustomer)I(Bbarber)J(Bbarber)


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Answers:

Dere is GoneG so!tion, "or which the initia a!es are m!tex & 5 (ariabe c!stomers may be

accessed), c!stomer & 9 (no c!stomers) and barber & 9 (barber is not b!sy).oidG c!stomer() {

H(Pm!tex)M

i" (c!stomers % ?) {K(Pm!tex)M

ret!rn ?*FFM

c!stomers R& 5M

K(Pm!tex)M

K(Pc!stomer)M

H(Pbarber)M

getDairC!t()M

H(Pm!tex)M

c!stomers & 5MK(Pm!tex)M

ret!rn ?*FFM

oidG barber() {

whie(5) {H(Pc!stomer)M

K(Pbarber)M

c!tDair()M


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Question 1': Synhroni!ation 2

Consider the "oowing three threads and semaphores

Knitiali#in/ semapores

s1 1

s2 9

s3 9

Knitiali#e x

x 9

void tread1() *

x x 1

void tread2() *

x x 2

void tread3() *

x x 2

Add H() and K() cas (and change nothing ese) to the three threads s!ch that at the end

o" any seE!ence o" conc!rrent exec!tion o" the three threads, x wi aways be & 6.


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Answer


s1 1

s2 9s3 9

Knitiali#e x

x 9

void tread1() * I(s1)

x x 1

J(s1)

J(s2)

void tread2() * I(s1)

x x 2

J(s1)J(s3)

void tread3() * I(s2) I(s3)

x x 2

J(s3)

J(s2)


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Question 11: Synhroni!ation 3

Consider the "oowing three threads and semaphores


s1

s2

s3

s4

Knitiali#e x

x 1

void tread1() * ile (x L 369) *

x x 2

exit(9)


x x 3

exit(9)


x x 5

exit(9)

Hroide initia a!es "or the "o!r semaphores and add H(), K() semaphore operations

(!sing the "o!r semaphores) in the code "or thread 5, 2 and s!ch that the process is

g!aranteed to terminate.


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Answer Knitiali#in/ semapores

s1 3

s2 2

s3 1s4 1

Knitiali#e x

x 1

void tread1() * ile (x L 369) * I(s1) I(s4) x x 2

J(s4) exit(9)


J(s4) exit(9)


J(s4) exit(9)

Final Review 09

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