EBK ACTIVITIES MANUAL FOR PROGRAMMABLE
EBK ACTIVITIES MANUAL FOR PROGRAMMABLE
5th Edition
ISBN: 8220102795983
Author: Petruzella
Publisher: YUZU
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Chapter 14, Problem 31RQ
Program Plan Intro

Access method:

  • The bus topology network needs some controlling techniques for access of particular devices to the bus.
  • For transmitting information, a Programmable Logic Controller (PLC) will access the network by using access method.
  • The access method on a network is capable of controlling the access of a particular device data to the bus.
  • The most common access methods include token passing, collision detection, and polling.

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3.) 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.
Returns 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.
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:

Chapter 14 Solutions

EBK ACTIVITIES MANUAL FOR PROGRAMMABLE

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