2024 – CMIS 102 Hands On Lab Week 5 Overview This hands on lab allows you to

 

CMIS 102 Hands-On Lab

// 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 %fn”,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:

1. What would you change in the design and the code if you wanted to input 10 students and 5 exams?
2. 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) ?

1. 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;