CMIS 102 HANDS-ON LAB WEEK 5.

CMIS 102 Hands-On Lab

// Week 5

This hands-on lab allows you to follow and experiment with the critical steps of developing a program including the program description, analysis, test plan, design (using pseudocode visualization), and implementation with C code. The example provided uses sequential, repetition statements and nested repetition statements.

This program will calculate the average of 3 exams for 5 students. The program will ask the user to enter 5 student names. For each of the students, the program will ask for 3 exam scores. The average exam score for each student will be calculated and printed.

I will use sequential and repetition programming statements.

I will define one String to store student name: StudentName.

I will define three Float numbers: Examvalue, Sum, Avg to store exam values the sum of the exams and the average of the exams.

The sum will be calculated by this formula:

Sum = Sum + Examvalue

For example, if the first value entered was 80.0 and second was 90.0 and the third exam was 100.0:

sum = sum + Examvalue = 0.0 + 80.0

sum = 80.0 + 90.0 = 170.0

sum = 170.0 + 100.0 = 270.0

Avg is then calculated as:

Avg = sum/3.0

For example 270.0/3.0 = 90.0

A nested repetition loop can be used to loop through each of the 5 students and each of the 3 exams:

For (students=0; students <5; students++)

For (exams=0;exams<3;exams++)

End For

End For

Sum values will need to be reset for each student to ensure only one student data is used for calculations each time.

To verify this program is working properly the input values could be used for testing:

Test Case

Input

Expected Output

1

Studentname=Chris

Examvalue1=80.0

Examvalue2=90.0

Examvalue3=100.0

Studentname=John

Examvalue1=70.0

Examvalue2=90.0

Examvalue3=80.0

Studentname=Sally

Examvalue1=100.0

Examvalue2=100.0

Examvalue3=100.0

Studentname=Pat

Examvalue1=50.0

Eexamvalue2=70.0

Examvalue3=60.0

Studentname=Sam

Examvalue1=90.0

Examvalue2=95.0

Examvalue3=100.0

Average for Chris is 90.0

Average for John is 80.0

Average for Sally is 100.0

Average for Pat is 60.0

Average for Sam is 95.0

// This program will calculate the average of 3 exams for 5 students

// Declare variables

Declare StudentName as String

Declare ExamValue, Sum, Avg as Float

// Loop through 5 Students

For (students=0; students <5 ; students++)

// reset Sum to 0

Set Sum =0.0

Print “Enter Student Name”

Input StudentName

// Nested Loop for Exams

For (exams=0; exams < 3; exams++)

Print “Enter exam grade: \n”

Input ExamValue

Set Sum = Sum + ExamValue

End For

Set Avg = Sum/3.0

Print “Average for “ + StudentName + “ is “ + Avg

End For

The following is the C Code that will compile in execute in the online compilers.

// C code

// This program will calculate the average of 3 exams for 5 students.

// Developer: Faculty CMIS102

// Date: Jan 31, 2014

#include <stdio.h>

int main ()

{

/* variable definition: */

char StudentName[100];

float ExamValue, Sum, Avg;

int students,exams;

// Loop through 5 Students

for (students=0; students <5 ; students++)

{

// reset Sum to 0

Sum =0.0;

printf(“Enter Student Name \n”);

scanf(“%s”, StudentName);

// Nested Loop for Exams

for (exams=0; exams < 3; exams++)

{

printf (“Enter exam grade: \n”);

scanf(“%f”, &ExamValue);

Sum = Sum + ExamValue;

}

Avg = Sum/3.0;

printf( “Average for %s is %f\n”,StudentName,Avg);

}

return 0;

}

Setting up the code and the input parameters in ideone.com:

Note the Student and ExamValues for this run were:

John: 90.0 80.0 100.0

Jim: 80.0 70.0 90.0

Joe: 70.0 100.0 100.0

Sally: 100.0 95.0 91.0

Sam: 30.0 54.0 68.0

You can change these values to any valid integer values to match your test cases.

Results from running the programming at ideone.com:

// Week 5

**Overview:**This hands-on lab allows you to follow and experiment with the critical steps of developing a program including the program description, analysis, test plan, design (using pseudocode visualization), and implementation with C code. The example provided uses sequential, repetition statements and nested repetition statements.

**Program Description:**This program will calculate the average of 3 exams for 5 students. The program will ask the user to enter 5 student names. For each of the students, the program will ask for 3 exam scores. The average exam score for each student will be calculated and printed.

**Analysis:**I will use sequential and repetition programming statements.

I will define one String to store student name: StudentName.

I will define three Float numbers: Examvalue, Sum, Avg to store exam values the sum of the exams and the average of the exams.

The sum will be calculated by this formula:

Sum = Sum + Examvalue

For example, if the first value entered was 80.0 and second was 90.0 and the third exam was 100.0:

sum = sum + Examvalue = 0.0 + 80.0

sum = 80.0 + 90.0 = 170.0

sum = 170.0 + 100.0 = 270.0

Avg is then calculated as:

Avg = sum/3.0

For example 270.0/3.0 = 90.0

A nested repetition loop can be used to loop through each of the 5 students and each of the 3 exams:

For (students=0; students <5; students++)

For (exams=0;exams<3;exams++)

End For

End For

Sum values will need to be reset for each student to ensure only one student data is used for calculations each time.

**Test Plan:**To verify this program is working properly the input values could be used for testing:

Test Case

Input

Expected Output

1

Studentname=Chris

Examvalue1=80.0

Examvalue2=90.0

Examvalue3=100.0

Studentname=John

Examvalue1=70.0

Examvalue2=90.0

Examvalue3=80.0

Studentname=Sally

Examvalue1=100.0

Examvalue2=100.0

Examvalue3=100.0

Studentname=Pat

Examvalue1=50.0

Eexamvalue2=70.0

Examvalue3=60.0

Studentname=Sam

Examvalue1=90.0

Examvalue2=95.0

Examvalue3=100.0

Average for Chris is 90.0

Average for John is 80.0

Average for Sally is 100.0

Average for Pat is 60.0

Average for Sam is 95.0

**Pseudocode:**// This program will calculate the average of 3 exams for 5 students

// Declare variables

Declare StudentName as String

Declare ExamValue, Sum, Avg as Float

// Loop through 5 Students

For (students=0; students <5 ; students++)

// reset Sum to 0

Set Sum =0.0

Print “Enter Student Name”

Input StudentName

// Nested Loop for Exams

For (exams=0; exams < 3; exams++)

Print “Enter exam grade: \n”

Input ExamValue

Set Sum = Sum + ExamValue

End For

Set Avg = Sum/3.0

Print “Average for “ + StudentName + “ is “ + Avg

End For

**Flow Chart:****C Code**The following is the C Code that will compile in execute in the online compilers.

// C code

// This program will calculate the average of 3 exams for 5 students.

// Developer: Faculty CMIS102

// Date: Jan 31, 2014

#include <stdio.h>

int main ()

{

/* variable definition: */

char StudentName[100];

float ExamValue, Sum, Avg;

int students,exams;

// Loop through 5 Students

for (students=0; students <5 ; students++)

{

// reset Sum to 0

Sum =0.0;

printf(“Enter Student Name \n”);

scanf(“%s”, StudentName);

// Nested Loop for Exams

for (exams=0; exams < 3; exams++)

{

printf (“Enter exam grade: \n”);

scanf(“%f”, &ExamValue);

Sum = Sum + ExamValue;

}

Avg = Sum/3.0;

printf( “Average for %s is %f\n”,StudentName,Avg);

}

return 0;

}

Setting up the code and the input parameters in ideone.com:

Note the Student and ExamValues for this run were:

John: 90.0 80.0 100.0

Jim: 80.0 70.0 90.0

Joe: 70.0 100.0 100.0

Sally: 100.0 95.0 91.0

Sam: 30.0 54.0 68.0

You can change these values to any valid integer values to match your test cases.

Results from running the programming at ideone.com:

**Learning Exercises for you to try:**- What would you change in the design and the code if you wanted to input 10 students and 5 exams?
- What is the line of code doing?

char StudentName[100];

(Hint: We haven’t covered arrays, but a String can be thought of as an array of characters) ?

- What would happen if you moved the Set Sum = 0.0 from inside the for loop to right after the declaration. For example:

// Declare variables

Declare StudentName as String

Declare ExamValue, Sum, Avg as Float

// Initialize Sum

Set Sum = 0.0;