C++ Program to Find Grade from Marks

Use a C++ program with if-else statements to check marks and print the corresponding grade, for example: if (marks >= 90) grade = 'A'; else if (marks >= 80) grade = 'B'; and so on.

Examples

Input95

OutputGrade: A

Input76

OutputGrade: C

Input45

OutputGrade: F

How to Think About It

To find the grade from marks, first understand the grading scale. Then compare the marks against the scale using if-else conditions starting from the highest grade to the lowest. Assign the grade based on the first condition that matches.

Algorithm

1

Get the marks input from the user.

2

Check if marks are 90 or above, assign grade A.

3

Else if marks are 80 or above, assign grade B.

4

Else if marks are 70 or above, assign grade C.

5

Else if marks are 60 or above, assign grade D.

6

Else assign grade F.

7

Print the grade.

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int marks;
    char grade;
    cout << "Enter marks: ";
    cin >> marks;

    if (marks >= 90) grade = 'A';
    else if (marks >= 80) grade = 'B';
    else if (marks >= 70) grade = 'C';
    else if (marks >= 60) grade = 'D';
    else grade = 'F';

    cout << "Grade: " << grade << endl;
    return 0;
}

Output

Enter marks: 85
Grade: B

Dry Run

Let's trace the program with input marks = 85 through the code

1

Input marks

User enters 85

2

Check if marks >= 90

85 >= 90 is false, move to next condition

3

Check if marks >= 80

85 >= 80 is true, assign grade = 'B'

4

Print grade

Output: Grade: B

Condition	Result	Grade Assigned
marks >= 90	false
marks >= 80	true	B

Why This Works

Step 1: Input marks

The program reads the marks entered by the user using cin.

Step 2: Check conditions

It uses if-else statements to compare marks starting from the highest grade boundary.

Step 3: Assign grade

Once a condition is true, the corresponding grade is assigned and the rest are skipped.

Step 4: Output grade

The program prints the grade using cout.

Alternative Approaches

Using switch with ranges (C++20 feature)

cpp

#include <iostream>
using namespace std;

int main() {
    int marks;
    char grade;
    cout << "Enter marks: ";
    cin >> marks;

    switch (marks / 10) {
        case 10:
        case 9: grade = 'A'; break;
        case 8: grade = 'B'; break;
        case 7: grade = 'C'; break;
        case 6: grade = 'D'; break;
        default: grade = 'F';
    }

    cout << "Grade: " << grade << endl;
    return 0;
}

This uses integer division and switch-case for cleaner code but only works well for marks 0-100.

Using ternary operator

cpp

#include <iostream>
using namespace std;

int main() {
    int marks;
    cout << "Enter marks: ";
    cin >> marks;

    char grade = (marks >= 90) ? 'A' :
                 (marks >= 80) ? 'B' :
                 (marks >= 70) ? 'C' :
                 (marks >= 60) ? 'D' : 'F';

    cout << "Grade: " << grade << endl;
    return 0;
}

This approach is concise but can be harder to read for beginners.

Complexity: O(1) time, O(1) space

Time Complexity

The program uses a fixed number of comparisons with no loops, so it runs in constant time O(1).

Space Complexity

Only a few variables are used, so space complexity is constant O(1).

Which Approach is Fastest?

All approaches run in constant time; switch-case may be slightly faster but less flexible.

Approach	Time	Space	Best For
if-else	O(1)	O(1)	Simple and clear grading logic
switch-case	O(1)	O(1)	Cleaner code for fixed ranges
ternary operator	O(1)	O(1)	Concise code, less readable