EBK BUSINESS DRIVEN TECHNOLOGY
7th Edition
ISBN: 8220103675451
Author: BALTZAN
Publisher: YUZU
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Chapter 13, Problem 1OC
Program Plan Intro
Disruptive technology:
This technology reanalyzes the competitive playing fields of their corresponding markets. Its motive is to destroy old markets and open new ones.
Sustaining technology:
The motive of this technology is to provide the customers with best, faster and inexpensive products in well-known markets.
Expert Solution & Answer
Explanation of Solution
Slack:
It is a software as a service application. This application is free to use and it can be connected with third-party applications like Drop box, Google drive, etc.
Slack is a sustaining technology:
Slack is a sustaining technology because of its simple IT structure that wraps various services. Sooner, e-mail communication could be replaced by slack because of its efficient working. Slack is a sustaining technology.
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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 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:
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--) {
In 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 ();
Chapter 13 Solutions
EBK BUSINESS DRIVEN TECHNOLOGY
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