Home » C# Creating a Grading Program

C# Creating a Grading Program

Solutons:


You are looking for else if construction:

  //TODO: out the right thresholds 
  static string GradeMe(int average) {
    if (average >= 90)           // 90+ 
      return "Grade A";
    else if (average >= 80)      // [80..90)
      return "Grade B";
    else if (average >= 70)      // [70..80)
      return "Grade C";
    else if (average >= 50)      // [50..70) 
      return "Grade D";
    else                         // less than 50
      return "Grade E";       
  }

Ternary operator ? : is a shorter but less readable alternative:

  static string GradeMe(int average) {
    return average >= 90 ? "Grade A"
         : average >= 80 ? "Grade B"
         : average >= 70 ? "Grade C"
         : average >= 50 ? "Grade D"
         : "Grade E";
  } 

Edit: If you want repeating, a loop is the solution; I suggest extracting method ReadGrade as well:

   static int ReadGrade(string title) {
     while (true) {
       Console.WriteLine(title);

       int result = 0;

       if (int.TryParse(Console.ReadLine(), out result) &&
           result >= 0 &&
           result <= 100)
         return result;

       Console.WriteLine("Please, try again"); 
     }
   }

   static void Main(string[] args) {
     do {
       int medeltal = (ReadGrade("math: ") +
                       ReadGrade("gym: ") + 
                       ReadGrade("classroom: ")) / 3;

       Console.WriteLine(GradeMe(medeltal));

       Console.WriteLine("Next grade (Y/N)?");
     }
     while (string.Equals("Y", 
                           Console.ReadLine().Trim(), 
                           StringComparison.OrdinalIgnoreCase))
   }

*first you do not defined grade32 but a grade2

*second they (grade1, grade2, grade32, …) are not the incoming values, that the caller method pass to the GradeMe() method, but the average so you should compare those value with your Literal values.

third, your first condition will trap item with score of 90 to 100, so not that you can’t, but you do not need to mention it again in your second condition expression, and just say it’s bigger than 75 …

if(x>=90 && x<=100) return "A";
if(x>=75) return "B";
if(x>=50) return "C";
....
return "F";

since you do use return, you also do not need the else, since the method will return, and you can omit all elses.

In function GradeMe() you create const int grades1 = 90; and const int grades2 = 75; and then check condition if (grades1 >= 90 && grades1 <= 100), which is always true, because grades1 >= 90 is true and grades1 <= 100 is true. So your else statement will never be reached.

You have to check average value in “if” =>

if (average >= 90 && average <= 100)
{
    return "Grade A";
}
else if (average >= 75 && average < 90)
{
    return "Grade B";
}
else
{
    return "Grade C";
}

and etc.

The most important thing is that all parts of function must return some value. You have to write ‘else’ for the last case

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