Programmable Logic Controllers
5th Edition
ISBN: 9780073373843
Author: Frank D. Petruzella
Publisher: McGraw-Hill Education
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Expert Solution & Answer
Chapter 15.2, Problem 10RQ
Explanation of Solution
Retentive output instruction:
- Output latch instruction (OTL) is considered to be a retentive output instruction and the meaning of retentive in the OTL is discussed below:
- Retentive is used to maintain or handle the status of the output.
- Retentive values or instructions keep their preceding state during a power cycle...
Expert Solution & Answer
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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 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:
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:
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,…
In 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…
Chapter 15 Solutions
Programmable Logic Controllers
Ch. 15.1 - Prob. 1RQCh. 15.1 - Prob. 2RQCh. 15.1 - Prob. 3RQCh. 15.1 - Prob. 4RQCh. 15.1 - Prob. 5RQCh. 15.1 - Prob. 6RQCh. 15.1 - Prob. 7RQCh. 15.1 - Prob. 8RQCh. 15.1 - Prob. 9RQCh. 15.1 - Prob. 10RQ
Ch. 15.1 - Prob. 11RQCh. 15.1 - Compare the accessibility of program scope and...Ch. 15.1 - Prob. 13RQCh. 15.1 - What is the difference between a produced tag and...Ch. 15.1 - Prob. 15RQCh. 15.1 - State the data type used for each of the...Ch. 15.1 - Describe the make-up of a predefined structure.Ch. 15.1 - Describe the make-up of a module-defined...Ch. 15.1 - Describe the make-up of a user-defined structure.Ch. 15.1 - Prob. 20RQCh. 15.1 - Prob. 21RQCh. 15.1 - Prob. 22RQCh. 15.1 - Prob. 23RQCh. 15.2 - Prob. 1RQCh. 15.2 - Prob. 2RQCh. 15.2 - Prob. 3RQCh. 15.2 - Prob. 4RQCh. 15.2 - Prob. 5RQCh. 15.2 - Prob. 6RQCh. 15.2 - Prob. 7RQCh. 15.2 - Prob. 8RQCh. 15.2 - Prob. 9RQCh. 15.2 - Prob. 10RQCh. 15.2 - Prob. 11RQCh. 15.2 - Extend control of the original ControlLogix...Ch. 15.2 - Prob. 3PCh. 15.3 - Prob. 1RQCh. 15.3 - Prob. 2RQCh. 15.3 - Prob. 3RQCh. 15.3 - Prob. 4RQCh. 15.3 - Prob. 5RQCh. 15.3 - Prob. 6RQCh. 15.3 - Prob. 7RQCh. 15.3 - Prob. 8RQCh. 15.3 - Prob. 9RQCh. 15.3 - Prob. 10RQCh. 15.3 - Prob. 11RQCh. 15.3 - Prob. 12RQCh. 15.3 - Modify the original CLX ten-second TON timer...Ch. 15.3 - Prob. 2PCh. 15.3 - Prob. 3PCh. 15.3 - Prob. 4PCh. 15.3 - Prob. 5PCh. 15.3 - Prob. 6PCh. 15.4 - Prob. 1RQCh. 15.4 - Prob. 2RQCh. 15.4 - Prob. 3RQCh. 15.4 - Prob. 4RQCh. 15.4 - Prob. 5RQCh. 15.4 - Prob. 6RQCh. 15.4 - Prob. 7RQCh. 15.4 - Prob. 1PCh. 15.4 - Prob. 2PCh. 15.5 - Prob. 1RQCh. 15.5 - Prob. 2RQCh. 15.5 - Prob. 3RQCh. 15.5 - Prob. 4RQCh. 15.5 - Prob. 5RQCh. 15.5 - Construct a ControlLogix ladder rung with compare...Ch. 15.5 - Prob. 2PCh. 15.5 - A single pole switch is used in place of the two...Ch. 15.6 - Prob. 1RQCh. 15.6 - Name the four basic elements of an FBD.Ch. 15.6 - Prob. 3RQCh. 15.6 - Prob. 4RQCh. 15.6 - Prob. 5RQCh. 15.6 - Prob. 6RQCh. 15.6 - Prob. 7RQCh. 15.6 - Prob. 8RQCh. 15.6 - Prob. 9RQCh. 15.6 - Prob. 10RQCh. 15.6 - Prob. 11RQCh. 15.6 - How is a function block feedback loop created?Ch. 15.6 - Prob. 13RQCh. 15.6 - Prob. 14RQCh. 15.6 - Prob. 1PCh. 15.6 - Prob. 2PCh. 15.6 - Prob. 3PCh. 15.6 - Prob. 4PCh. 15.6 - Prob. 5P
Knowledge Booster
Similar questions
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