INFORMATION TECH.PROJECT...-W/MINDTAP
INFORMATION TECH.PROJECT...-W/MINDTAP
9th Edition
ISBN: 9781337586801
Author: SCHWALBE
Publisher: CENGAGE L
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Chapter 6, Problem 6DQ

Explanation of Solution

Gantt Charts:

A Gantt chart also known as bar chart. It is used for displaying the schedule information by specifying the activities of the project and the dates like when they start and finish in a specific calendar format.

  • Gantt chart for the software projects usually contains milestones, tasks, task durations and arrows showing task dependencies.
  • In software project Gantt chart, we generally have a black diamond symbol, thick black bars, grey horizontal bars, arrows and white diamond symbols.
  • The black diamond symbol generally represents a milestone.
  • The thick black bars with arrows represent the summary tasks.
  • The grey horizontal bars represent the duration of each task.
  • Arrows show the relationship between the tasks.
  • The white diamond represents a missed milestone.

Critical Path Method:

Critical Path Method is a network diagramming technique. It is used for predicting the duration of the total project.

  • This tool helps in combating the schedule overruns.
  • It is also known as critical path analysis.
  • Critical path is nothing but a series of activities, which determines the earliest time that a project needs to be completed.
  • It has a less amount of slack and float.
  • Slack and float is the time in which an activity can be delayed without actually delaying the project finish project.
  • In order to develop a critical path for a project, we need to develop a network diagram first.
  • Once the network diagram is created we can determine the critical path by estimating the duration of each activity.
  • We use critical path to make schedule Trade-offs and to shorten a project schedule...

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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?…
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