Fundamentals of Information Systems (Looseleaf)
9th Edition
ISBN: 9781337099042
Author: STAIR
Publisher: Cengage
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Question
Chapter 7, Problem 19RQ
Program Plan Intro
Difference between file conversion and file compression.
Expert Solution & Answer
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1234
3. Which line prevents compiler optimization? Circle one: 1234
Suggested solution:
Store strlen(str) in a variable before the if statement.
⚫ Remove the if statement.
Replace index 0 && index < strlen(str)) {
5 }
}
=
str [index] = val;
Character Hex value | Character Hex value Character Hex value
'A'
0x41
'J'
Ox4a
'S'
0x53
'B'
0x42
'K'
0x4b
"T"
0x54
0x43
'L'
Ox4c
'U'
0x55
0x44
'M'
0x4d
'V'
0x56
0x45
'N'
Ox4e
'W'
0x57
0x46
'0'
Ox4f
'X'
0x58
0x47
'P'
0x50
'Y'
0x59
0x48
'Q'
0x51
'Z'
Ox5a
'T'
0x49
'R'
0x52
'\0'
0x00
Now consider what happens on a Linux/x86 machine when callfoo calls foo with
the input string "ZYXWVUTSRQPONMLKJIHGFEDCBA".
A. On the left draw the state of the stack just before the execution of the instruction
at address Ox40053a; make sure to show the frames for callfoo and foo and
the exact return address, in Hex at the bottom of the callfoo frame.
Then, on the right, draw the state of the stack just after the instruction got
executed; make sure to show where the string "ZYXWVUTSRQPONMLKJIHGFEDCBA"
is placed and what part, if any, of the above return address has been overwritten.
B. Immediately after the ret instruction at address 0x400543 executes, what is
the value of the program counter register %rip?…
1 typedef struct node* {
2
struct node* next;
3
char* key;
4
char* val;
5} node_t;
6
7 char* find_node (node_t* node, char* key_to_find) {
while(strcmp (node->key, key_to_find ) != 0 ) {
node = node->next;
8
9
10
}
11
return node->val;
12 }
Chapter 7 Solutions
Fundamentals of Information Systems (Looseleaf)
Ch. 7 - Prob. 1LOCh. 7 - Prob. 2LOCh. 7 - Prob. 3LOCh. 7 - Prob. 4LOCh. 7 - Prob. 5LOCh. 7 - Prob. 6LOCh. 7 - Prob. 7LOCh. 7 - Prob. 8LOCh. 7 - Prob. 9LOCh. 7 - Prob. 1.1RQ
Ch. 7 - Prob. 1.2RQCh. 7 - Prob. 1.1CTQCh. 7 - Prob. 1.2CTQCh. 7 - Prob. 2.1RQCh. 7 - Prob. 2.2RQCh. 7 - Prob. 2.1CTQCh. 7 - Prob. 2.2CTQCh. 7 - Prob. 3.1RQCh. 7 - Prob. 3.2RQCh. 7 - Prob. 3.1CTQCh. 7 - Prob. 3.2CTQCh. 7 - What sort of training and experience is needed by...Ch. 7 - Prob. 4.2RQCh. 7 - Prob. 4.1CTQCh. 7 - Prob. 4.2CTQCh. 7 - Prob. 1SATCh. 7 - Prob. 2SATCh. 7 - Prob. 3SATCh. 7 - Prob. 4SATCh. 7 - Prob. 5SATCh. 7 - Prob. 6SATCh. 7 - Prob. 7SATCh. 7 - Prob. 8SATCh. 7 - Prob. 9SATCh. 7 - Prob. 10SATCh. 7 - Prob. 11SATCh. 7 - Prob. 12SATCh. 7 - Prob. 13SATCh. 7 - Prob. 14SATCh. 7 - Prob. 15SATCh. 7 - Prob. 16SATCh. 7 - Prob. 17SATCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQCh. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 16RQCh. 7 - Prob. 17RQCh. 7 - Prob. 18RQCh. 7 - Prob. 19RQCh. 7 - Prob. 20RQCh. 7 - Prob. 21RQCh. 7 - Prob. 22RQCh. 7 - Prob. 23RQCh. 7 - Prob. 1DQCh. 7 - Prob. 2DQCh. 7 - Prob. 3DQCh. 7 - Prob. 4DQCh. 7 - Prob. 5DQCh. 7 - Prob. 6DQCh. 7 - Prob. 7DQCh. 7 - Prob. 8DQCh. 7 - Prob. 9DQCh. 7 - Prob. 10DQCh. 7 - Prob. 11DQCh. 7 - Prob. 12DQCh. 7 - Prob. 13DQCh. 7 - Prob. 1PSECh. 7 - Prob. 2PSECh. 7 - Prob. 3PSECh. 7 - Prob. 1WECh. 7 - Prob. 2WECh. 7 - Prob. 3WECh. 7 - Prob. 1CECh. 7 - Prob. 2CECh. 7 - Prob. 3CECh. 7 - Prob. 1.1CSCh. 7 - Prob. 1.2CSCh. 7 - Prob. 1.3CSCh. 7 - Prob. 1.4CSCh. 7 - Prob. 1.5CSCh. 7 - Prob. 2.1CSCh. 7 - Prob. 2.2CSCh. 7 - Prob. 2.3CS
Knowledge Booster
Similar questions
- Match each of the assembler routines on the left with the equivalent C function on the right. Write the name of the label (e.g., foo) to the right of the corresponding function. Note: shrq is the logical right shift instruction, and sarq is the arithmetic right shift instruction. foo1: leaq 0(,%rdi, 8), %rax long choice1 (long x) { ret return x - 8 >8; foo3: } movq sarq %rdi, %rax $8, %rax long choice4 (long x) ret { return x*256; } foo4: long choice5 (long x) leaq -8 (%rdi), %rax { ret return x-8; } long choice6 (long x) foo5: { leaq -8 (%rdi), %rax return x+8; shrq $63, %rax } retarrow_forwardGiven the variables and code in the text below, identify where in memory they will live once the code is compiled. 1 char big_array [1L<<24]; /* 16 MB */ 2 GB * :/ 2 char huge_array [1L<<31]; /* 3 4 int global = 0; 5 6 int useless () { return 0; } 7 8 int main() 9 { 10 void *p1, p2, *p3, *p4; int local = 0; malloc (1L << 28); /* 256 MB *, 11 12 p1 13 p2 = malloc (1L << 8); /* 256 B * 14 p3 15 p4 = malloc (1L << 32); malloc (1L << 8); /* 4 GB * */ /* 256 B */ 16 } Note: *pN is the thing at which pN points. 1. big_array 2. huge_array 3. global 4. useless 5. void* p1 6. *p1 7. void* p2 8. *p2 9. void* p3 10. *p3 11. void* p4 12. *p4arrow_forwardThe next problem concerns the following C code: /copy input string x to buf */ void foo (char *x) { char buf [8]; strcpy((char *) buf, x); } void callfoo() { } foo("ZYXWVUTSRQPONMLKJIHGFEDCBA"); Here is the corresponding machine code on a Linux/x86 machine: 0000000000400530 : 400530: 48 83 ec 18 sub $0x18,%rsp 400534: 48 89 fe mov %rdi, %rsi 400537: 48 89 e7 mov %rsp,%rdi 40053a: e8 di fe ff ff callq 400410 40053f: 48 83 c4 18 add $0x18,%rsp 400543: c3 retq 400544: 0000000000400544 : 48 83 ec 08 sub $0x8,%rsp 400548: bf 00 06 40 00 mov $0x400600,%edi 40054d: e8 de ff ff ff callq 400530 400552: 48 83 c4 08 add $0x8,%rsp 400556: c3 This problem tests your understanding of the program stack. Here are some notes to help you work the problem: ⚫ strcpy(char *dst, char *src) copies the string at address src (including the terminating '\0' character) to address dst. It does not check the size of the destination buffer. • You will need to know the hex values of the following characters:arrow_forward
- Consider the following assembly code for a C for loop: movl $0, %eax jmp .L2 .L3: addq $1, %rdi addq %rsi, %rax subq $1, %rsi .L2: cmpq %rsi, %rdi jl .L3 addq ret %rdi, %rax Based on the assembly code above, fill in the blanks below in its corresponding C source code. Recall that registers %rdi and %rsi contain the first and second, respectively, argument of a function. (Note: you may only use the symbolic variables x, y, and result in your expressions below do not use register names.) long loop (long x, long y) { long result; } for ( } return result; __; y--) {arrow_forwardIn each of the following C code snippets, there are issues that can prevent the compiler from applying certain optimizations. For each snippet: Circle the line number that contains compiler optimization blocker. ⚫ Select the best modification to improve optimization. 1. Which line prevents compiler optimization? Circle one: 2 3 4 5 6 Suggested solution: ⚫ Remove printf or move it outside the loop. Remove the loop. • Replace arr[i] with a constant value. 1 int sum (int *arr, int n) { 2 int s = 0; 3 for (int i = 0; i < n; i++) { 4 5 6 } 7 8 } s = arr[i]; printf("%d\n", s); return s; 234206 2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6 Suggested solution: Move or eliminate do_extra_work() if it's not necessary inside the loop. Remove the loop (but what about scaling?). ⚫ Replace arr[i] *= factor; with arr[i] = 0; (why would that help?). 1 void scale (int *arr, int n, int factor) { 5 6 } for (int i = 0; i < n; i++) { rr[i] = factor; do_extra_work ();arrow_forward123456 A ROP (Return-Oriented Programming) attack can be used to execute arbitrary instructions by chaining together small pieces of code called "gadgets." Your goal is to create a stack layout for a ROP attack that calls a function located at '0x4018bd3'. Below is the assembly code for the function 'getbuf', which allocates 8 bytes of stack space for a 'char' array. This array is then passed to the 'gets' function. Additionally, you are provided with five useful gadgets and their addresses. Use these gadgets to construct the stack layout. Assembly for getbuf 1 getbuf: sub mov $8, %rsp %rsp, %rdi call gets add $8, %rsp ret #Allocate 8 bytes for buffer #Load buffer address into %rdi #Call gets with buffer #Restore the stack pointer #Return to caller Stack Layout each 8-byte (fill in section) Address Value (8 bytes) 0x7fffffffdfc0 0x7fffffffdfb8 0x7fffffffdfb0 0x7fffffffdfa8 0x7fffffffdfa0 0x7fffffffdf98 0x7fffffffdf90 0x7fffffffdf88 Gadgets Address Gadget Ox4006a7 pop %rdi; ret Ox4006a9…arrow_forward
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