Answered: In Programming Exercise 13 (Chapter 8),…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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In Programming Exercise 13 (Chapter 8), you are asked to write a program to calculate students’ average test scores and their grades. Improve this programming exercise by adding a function to sort students’ names so that students’ data is output into ascending order according to their name.

(data.txt)

Johnson 85 83 77 91 76

Cooper 78 81 11 90 73

Gupta 92 83 30 69 87

Blair 23 45 96 38 59

Clark 60 85 45 39 67

Kennedy 77 31 52 74 83

Bronson 93 94 89 77 97

Sunny 79 85 28 93 82

Smith 85 72 49 75 63

Aniston 80 90 95 93 48

(Program to Improve)

#include <iostream>

#include <fstream>

#include <string>

#include <iomanip>

using namespace std;

void getData(ifstream& inf, string n[], double tstData[][6], int count);

void calculateAverage(double tstData[][6], int count);

void calculateGrade(double tstData[][6], char gr[], int count);

void print(string n[], double tstData[][6], char gr[], int count);

int main()

{

string names[10];

double testData[10][6];

char grade[10];

ifstream inFile;

inFile.open("data.txt");

if (!inFile)

{

cout << "Cannot open the input file: ch8_Ex13Data.txt." << endl;

cout << "Program terminates!" << endl;

return 1;

}

cout << fixed << showpoint << setprecision(2);

getData(inFile, names, testData, 10);

calculateAverage(testData, 10);

calculateGrade(testData, grade, 10);

print(names, testData, grade, 10);

inFile.close();

return 0;

}

void getData(ifstream& inf, string n[], double tstData[][6], int count)

{

for (int i = 0; i < count; i++)

{

inf >> n[i];

for (int j = 0; j < 5; j++)

inf >> tstData[i][j];

tstData[i][5] = 0.0;

}

void calculateAverage(double tstData[][6], int count)

{

double sum;

for (int i = 0; i < count; i++)

{

sum = 0.0;

for (int j = 0; j < 5; j++)

sum = sum + tstData[i][j];

tstData[i][5] = sum / 5;

}

void calculateGrade(double tstData[][6], char gr[], int count)

{

for (int i = 0; i < count; i++)

if (tstData[i][5] >= 90)

gr[i] = 'A';

else if (tstData[i][5] >= 80)

gr[i] = 'B';

else if (tstData[i][5] >= 70)

gr[i] = 'C';

else if (tstData[i][5] >= 60)

gr[i] = 'D';

else

gr[i] = 'F';

}

void print(string n[], double tstData[][6], char gr[], int count)

{

double sum = 0.0;

cout << left << setw(10) << "Name"

<< right << setw(8) << "Test 1"

<< setw(8) << "Test 2"

<< setw(8) << "Test 3"

<< setw(8) << "Test 4"

<< setw(8) << "Test 5"

<< setw(10) << "Average"

<< setw(8) << "Grade" << endl;

for (int i = 0; i < count; i++)

{

cout << left << setw(10) << n[i];

cout << right;

for (int j = 0; j < 5; j++)

cout << setw(8) << tstData[i][j];

cout << setw(10) << tstData[i][5]

<< setw(6) << gr[i] << endl;

sum = sum + tstData[i][5];

}

cout << endl << endl;

cout << "Class average: " << sum / count << endl;

}

The Output must match all data below, say for the exception that it must be sorted. The last answer I received was less than satisfactory.

Name
Test 1 Test 2 Test 3 Test 4 Test 5
Average
Grade
Johnson
85.00
83.00
77.00
91.00
76.00
82.40
Сооper
Gupta
78.00
81.00
11.00
90.00
73.00
66.60
92.00
83.00
30.00
69.00
87.00
72.20
Blair
23.00
45.00
96.00
38.00
59.00
52.20
F
Clark
60.00
85.00
45.00
39.00
67.00
59.20
F
Kennedy
77.00
31.00
52.00
74.00
83.00
63.40
Bronson
93.00
94.00
89.00
77.00
97.00
90.00
A
Sunny
79.00
85.00
28.00
93.00
82.00
73.40
C
Smith
85.00
72.00
49.00
75.00
63.00
68.80
Aniston
80.00
90.00
95.00
93.00
48.00
81.20
Class average: 70.94

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