Computer Science Illuminated
6th Edition
ISBN: 9781284055917
Author: Nell Dale, John Lewis
Publisher: Jones & Bartlett Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 7, Problem 3TQ
Program Plan Intro
Problem Solving Strategies:
Problem solving is defined as the process of working on the description of a problem to obtain a solution.
Two types of problem solving strategy:
- Top-Down Approach:
- Large program is subdivided into small subprograms called modules.
- These modules are processed individually.
- If combining these modules then it produces the entire program.
- It is also called as Stepwise design.
- Bottom-Up Approach:
- In this approach, first the individual elements are specified in very detail.
- These individual elements are combined together which in turn combined in sequence of levels up to the complete larger system is obtained.
Explanation of Solution
Other approach used in problem solving strategy:
Bottom-Up Approach:
- In this approach, first the individual elements are specified in very detail...
Explanation of Solution
“Yes”, Bottom-up approach will be appropriate for problem solving.
Reasons:
- When existing components are taken into the account, bottom-up approach is used...
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The 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:
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?…
Chapter 7 Solutions
Computer Science Illuminated
Ch. 7 - Prob. 1ECh. 7 - Prob. 2ECh. 7 - Prob. 3ECh. 7 - Prob. 4ECh. 7 - Prob. 5ECh. 7 - Prob. 6ECh. 7 - Prob. 7ECh. 7 - Prob. 8ECh. 7 - Prob. 9ECh. 7 - Prob. 10E
Ch. 7 - Prob. 11ECh. 7 - Prob. 12ECh. 7 - Prob. 13ECh. 7 - Prob. 14ECh. 7 - Prob. 15ECh. 7 - Prob. 16ECh. 7 - Prob. 17ECh. 7 - Prob. 18ECh. 7 - Prob. 19ECh. 7 - Prob. 20ECh. 7 - Prob. 21ECh. 7 - Prob. 22ECh. 7 - Prob. 23ECh. 7 - Prob. 24ECh. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27ECh. 7 - Prob. 28ECh. 7 - Prob. 29ECh. 7 - Prob. 30ECh. 7 - Prob. 31ECh. 7 - Prob. 32ECh. 7 - Prob. 33ECh. 7 - Prob. 34ECh. 7 - Prob. 35ECh. 7 - Prob. 36ECh. 7 - Prob. 37ECh. 7 - Prob. 38ECh. 7 - Prob. 39ECh. 7 - Prob. 40ECh. 7 - Prob. 41ECh. 7 - Prob. 42ECh. 7 - Prob. 43ECh. 7 - Prob. 44ECh. 7 - Prob. 45ECh. 7 - Prob. 46ECh. 7 - Prob. 47ECh. 7 - Prob. 48ECh. 7 - Prob. 49ECh. 7 - Prob. 50ECh. 7 - Prob. 51ECh. 7 - Prob. 52ECh. 7 - Prob. 53ECh. 7 - Prob. 54ECh. 7 - Prob. 55ECh. 7 - Prob. 56ECh. 7 - Prob. 57ECh. 7 - Prob. 58ECh. 7 - Prob. 59ECh. 7 - Prob. 60ECh. 7 - Prob. 61ECh. 7 - Prob. 62ECh. 7 - Prob. 63ECh. 7 - Prob. 65ECh. 7 - Prob. 1TQCh. 7 - Prob. 2TQCh. 7 - Prob. 3TQCh. 7 - Prob. 4TQCh. 7 - Prob. 5TQCh. 7 - Prob. 6TQCh. 7 - Prob. 7TQCh. 7 - Prob. 8TQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.Similar questions
- 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 }arrow_forwardMatch 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_forward
- The 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_forwardConsider 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_forward
- 123456 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_forwardProblem 1 [15 points] The code below is buggy. Assume the code compiles. Briefly: 1). Identify the problem with the code (e.g., can access memory out of bounds) and 2). Suggest a solution (e.g., check the length). Question 1 1 #define BLENGTH 5 2 int b[BLENGTH]; 3 void copy_from_global_int_array_b (int n, int* dest) { 4 5 } *dest = b[n]; ==arrow_forwardWhich statement regarding SGA_MAX_SIZE is true? SGA_MAX_SIZE is modifiable after an instance is started, only when Automatic Memory Management is used. SGA_MAX_SIZE is not dyamically modifiable. SGA_MAX_SIZE is ignored when MEMORY_TARGET > 0. SGA-MAX_SIZE must be specified when SGA_TARGET > 0arrow_forward
- Explian this C program #include <stdio.h> unsigned int rotateRight(unsigned int num, unsigned int bits) { unsignedint bit_count =sizeof(unsignedint) *8; bits = bits % bit_count; // Handle cases where bits >= bit_count return (num >> bits) | (num << (bit_count - bits)); } int main() { unsignedint num, bits; printf("Enter a number: "); scanf("%u", &num); printf("Enter the number of bits to shift: "); scanf("%u", &bits); printf("After rotation: %u\n", rotateRight(num, bits)); return0; }arrow_forwardExplian thiS C program #include<stdio.h> int countSetBits(int n) { int count = 0; while (n) { count += n & 1; n >>= 1; } return count;} int main() { int num; printf("Enter a number: "); scanf("%d", &num); printf("Output: %d units\n", countSetBits(num)); return 0;}arrow_forwardPlease provide the Mathematica codearrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Database System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill Education
Starting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSON
Digital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSON
C How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSON
Database Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage Learning
Programmable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education
Database System Concepts
Computer Science
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:McGraw-Hill Education
Starting Out with Python (4th Edition)
Computer Science
ISBN:9780134444321
Author:Tony Gaddis
Publisher:PEARSON
Digital Fundamentals (11th Edition)
Computer Science
ISBN:9780132737968
Author:Thomas L. Floyd
Publisher:PEARSON
C How to Program (8th Edition)
Computer Science
ISBN:9780133976892
Author:Paul J. Deitel, Harvey Deitel
Publisher:PEARSON
Database Systems: Design, Implementation, & Manag...
Computer Science
ISBN:9781337627900
Author:Carlos Coronel, Steven Morris
Publisher:Cengage Learning
Programmable Logic Controllers
Computer Science
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education