Computing Essentials 2019 27th Edition
19th Edition
ISBN: 9781260096057
Author: By Timothy O'Leary and Linda O'Leary and Daniel O'Leary
Publisher: McGraw-Hill Education
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Chapter 13, Problem 24CC
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What distinguishes a lower-level language from a higher-level language?
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Chapter 13 Solutions
Computing Essentials 2019 27th Edition
Ch. 13 - Prob. 1CCCh. 13 - Prob. 2CCCh. 13 - Prob. 3CCCh. 13 - Prob. 4CCCh. 13 - Prob. 5CCCh. 13 - Prob. 6CCCh. 13 - Prob. 7CCCh. 13 - Prob. 8CCCh. 13 - Prob. 9CCCh. 13 - Prob. 10CC
Ch. 13 - Prob. 11CCCh. 13 - Prob. 12CCCh. 13 - Prob. 13CCCh. 13 - Prob. 14CCCh. 13 - Prob. 15CCCh. 13 - Prob. 16CCCh. 13 - Prob. 17CCCh. 13 - Prob. 18CCCh. 13 - Prob. 19CCCh. 13 - Prob. 20CCCh. 13 - Prob. 21CCCh. 13 - Prob. 22CCCh. 13 - Prob. 23CCCh. 13 - Prob. 24CCCh. 13 - Prob. 25CCCh. 13 - Prob. 26CCCh. 13 - Prob. 27CCCh. 13 - Prob. 1MCCh. 13 - Circle the correct answer. 2. The major processing...Ch. 13 - Prob. 3MCCh. 13 - Prob. 4MCCh. 13 - Prob. 5MCCh. 13 - Prob. 6MCCh. 13 - Circle the correct answer. 7. Unlike traditional...Ch. 13 - Prob. 8MCCh. 13 - Prob. 9MCCh. 13 - Prob. 10MCCh. 13 - Prob. 1MCh. 13 - Prob. 1OECh. 13 - Prob. 2OECh. 13 - Prob. 3OECh. 13 - Prob. 4OECh. 13 - Prob. 5OECh. 13 - Prob. 1DCh. 13 - Prob. 2aDCh. 13 - Prob. 2bD
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- The following table shows the timestamp and actions by two users. Choose the best option that describes the outcome of the actions. Time JohnSara 10:14 select* from hr.employees; 10:15 Update hr.employees set salary= 100 where employee_id= 206; 10:16 Commit: Select* from hr.employees; 10:18 Commit: 10:20 Select* from hr.employees; Commit: John's query willreturn the same results all three times it is executed as they are run in the same session. John's queries run at10:16 and10:20 produce the same result, which is different from the one at 10:14 John's query run at 10:16 waits until 10:18 to produce results, waiting for the commit to happen. John's queries run at 10:14 and 10:16 produce the same result, which is different from the one at 10:20arrow_forwardwhat's the process used to obtain IP configuration using DHCP in Windows Server.arrow_forwardConsider the following sequential circuit: CLOCK a. Define the diagram circuit variables (5 pts) b. Derive the Flip-Flop input equations) (5 pts) c. Derive the circuit output equation (5 pts) d. Derive the state table of the circuit (5 pts) e. Derive the state diagram for this circuit (5 pts) Clk A D B B' CIK Question 2 (25 pts) A sequential circuit with two D flip-flops A and B, two inputs x and y, and one output z is specified by the following next-state and output equations: A(t + 1) = xy' + xB B(t + 1) = xA + xB' z = A a. Draw the logic diagram of the circuit. (5 pts) b. List the state table for the sequential circuit. (10 pts) c. Draw the corresponding state diagram. (10 pts)arrow_forward
- 5. Word FrequencyWrite a program that reads the contents of a text file. The program should create a dictio-nary in which the keys are the individual words found in the file and the values are the number of times each word appears. For example, if the word “the” appears 128 times, the dictionary would contain an element with 'the' as the key and 128 as the value. The program should either display the frequency of each word or create a second file containing a list of each word and its frequency.arrow_forward3.) File Encryption and DecryptionWrite a program that uses a dictionary to assign “codes” to each letter of the alphabet. For example: codes = { ‘A’ : ‘%’, ‘a’ : ‘9’, ‘B’ : ‘@’, ‘b’ : ‘#’, etc . . .}Using this example, the letter A would be assigned the symbol %, the letter a would be assigned the number 9, the letter B would be assigned the symbol @, and so forth. The program should open a specified text file, read its contents, then use the dictionary to write an encrypted version of the file’s contents to a second file. Each character in the second file should contain the code for the corresponding character in the first file. Write a second program that opens an encrypted file and displays its decrypted contents on the screen.arrow_forwardReturns an US standard formatted phone number, in the format of (xxx) xxx-xxxx the AreaCode, Prefix and number being each part in order. Testing Hint: We be exact on the format of the number when testing this method. Make sure you think about how to convert 33 to 033 or numbers like that when setting your string format. Reminder the %02d - requires the length to be 2, with 0 padding at the front if a single digit number is passed in.arrow_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 d1 fe ff ff 40053f: 48 83 c4 18 add callq 400410 $0x18,%rsp 400543: c3 retq 0000000000400544 : 400544: 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_forwardA 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: 2 sub $8, %rsp 3 mov %rsp, %rdi 4 call gets 56 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 (fill in Gadgets each 8-byte section) Address Gadget Address Value (8 bytes) 0x4006a7 pop %rdi; ret 0x7fffffffdfc0 Ox4006a9 pop %rsi; ret 0x7fffffffdfb8 0x4006ab pop %rax; ret 0x7fffffffdfb0 0x7fffffffdfa8 Ox4006ad mov %rax,…arrow_forwardIn each of the following C code snippets, there are issues that can prevent the compilerfrom 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 6Suggested solution:• Remove printf or move it outside the loop.• Remove the loop.• Replace arr[i] with a constant value.1 int sum( int ∗ ar r , int n) {2 int s = 0 ;3 for ( int i = 0 ; i < n ; i++) {4 s += a r r [ i ] ;5 p r i n t f ( ”%d\n” , s ) ;6 }7 return s ;8 }2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6Suggested 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 s c a l e ( int ∗ ar r , int n , int f a c t o r ) {2 for ( int i = 0 ; i < n ; i++) {3 a r r [ i ] ∗= f a c t o r…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 'Ox4018bd3'. 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 6 ret #Allocate 8 bytes for buffer #Load buffer address into %rdi #Call gets with buffer #Restore the stack pointer #Return to caller Stack each Layout (fill in Gadgets 8-byte section) Address Gadget Address Value (8 bytes) 0x7fffffffdfc0 0x7fffffffdfb8 0x7fffffffdfb0 0x7fffffffdfa8 0x7fffffffdfa0 0x7fffffffdf98 0x7fffffffdf90 0x7fffffffdf88 Original 0x4006a7 pop %rdi;…arrow_forwardCharacter Hex value || Character Hex value | Character Hex value 'A' 0x41 יני Ox4a 'S' 0x53 0x42 'K' 0x4b 'T" 0x54 0x43 'L' Ox4c 0x55 0x44 'M' Ox4d 0x56 0x45 'N' Ox4e 'W' 0x57 0x46 Ox4f 'X' 0x58 0x47 'P' 0x50 'Y' 0x59 'H' 0x48 'Q' 0x51 'Z' Охба '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? (That is…arrow_forwardDraw out the way each of these structs looks in memory, including padding! Number the offsets in memory. 1 struct okay Name 2 { short a; 3 4 long number; 5 int also_a_number; 6 7 }; char* text; 1 struct badName 2 { 3 4 5 }; short s; struct okay Name n;arrow_forward
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