Three-subject results using one-dimensional arrays

LAB A06 · 01

Understand the task.

Aim: Three-subject results using one-dimensional arrays.

Review the related theory in Unit 3.

LAB A06 · 02

Know what goes in and what comes out.

First n, a student count from 1 to 20. Then n groups of three integer subject marks, each 0–100. Practice result rule: PASS only when all three marks are at least PASS_MARK, set to 40.

3
80 70 90
40 40 40
95 35 90

Student 1: total = 240, average = 80.00, result = PASS
Student 2: total = 120, average = 40.00, result = PASS
Student 3: total = 220, average = 73.33, result = FAIL

Source programs read the documented input directly, without prompt text, so the same sample works in the embedded compiler’s STDIN field.

LAB A06 · 03

Plan the steps before writing code.

1. Read and validate n.
2. Store the three marks for student i in first[i], second[i] and third[i].
3. For each student, total the three corresponding elements.
4. Compute average = total / 3.0.
5. Check every subject against PASS_MARK and print total, average and result.

LAB A06 · 04

One complete program. Ready to run.

#include <stdio.h>
#define MAX_STUDENTS 20
#define PASS_MARK 40
int main(void)
{
    int first[MAX_STUDENTS], second[MAX_STUDENTS], third[MAX_STUDENTS];
    int n, i, total, passed;
    double average;
    if (scanf("%d", &n) != 1 || n < 1 || n > MAX_STUDENTS) {
        puts("Enter a student count from 1 to 20.");
        return 1;
    }
    for (i = 0; i < n; i++) {
        if (scanf("%d%d%d", &first[i], &second[i], &third[i]) != 3 ||
            first[i] < 0 || first[i] > 100 || second[i] < 0 || second[i] > 100 ||
            third[i] < 0 || third[i] > 100) {
            puts("Enter three marks from 0 to 100 for each student.");
            return 1;
        }
    }
    for (i = 0; i < n; i++) {
        total = first[i] + second[i] + third[i];
        average = total / 3.0;
        passed = first[i] >= PASS_MARK && second[i] >= PASS_MARK && third[i] >= PASS_MARK;
        printf("Student %d: total = %d, average = %.2f, result = %s\n",
               i + 1, total, average, passed ? "PASS" : "FAIL");
    }
    return 0;
}

gcc A06-student-results.c -o program

LAB A06 · 05

Connect each statement to its purpose.

• Three separate 1D arrays meet the prescribed one-dimensional-array requirement. A marks[n][3] declaration would instead be a 2D array.
• The same index i always refers to the same student across the three arrays.
• Using 3.0 produces a fractional average.
• A high average does not compensate for a failed subject under this stated practice rule. Change PASS_MARK and the condition if your prescribed policy differs.

LAB A06 · 06

Follow the sample calculation.

LAB A06 · 07

Check more than one input.

Time and storage

O(n) time. Three arrays hold O(n) logical data, with a fixed capacity of 20 students in this program.

Start with the sample, then test a boundary and a case that changes the result. A correct sample alone does not establish that every input is handled correctly.

LAB A06 · 08

Use the real C compiler.

Choose an example and load it into the embedded C compiler. Select I/O inside the compiler, check the sample STDIN, and press Run. Change the input or edit the source to explore a different case.

3
80 70 90
40 40 40
95 35 90

Student 1: total = 240, average = 80.00, result = PASS
Student 2: total = 120, average = 40.00, result = PASS
Student 3: total = 220, average = 73.33, result = FAIL

#include <stdio.h>
#define MAX_STUDENTS 20
#define PASS_MARK 40
int main(void)
{
    int first[MAX_STUDENTS], second[MAX_STUDENTS], third[MAX_STUDENTS];
    int n, i, total, passed;
    double average;
    if (scanf("%d", &n) != 1 || n < 1 || n > MAX_STUDENTS) {
        puts("Enter a student count from 1 to 20.");
        return 1;
    }
    for (i = 0; i < n; i++) {
        if (scanf("%d%d%d", &first[i], &second[i], &third[i]) != 3 ||
            first[i] < 0 || first[i] > 100 || second[i] < 0 || second[i] > 100 ||
            third[i] < 0 || third[i] > 100) {
            puts("Enter three marks from 0 to 100 for each student.");
            return 1;
        }
    }
    for (i = 0; i < n; i++) {
        total = first[i] + second[i] + third[i];
        average = total / 3.0;
        passed = first[i] >= PASS_MARK && second[i] >= PASS_MARK && third[i] >= PASS_MARK;
        printf("Student %d: total = %d, average = %.2f, result = %s\n",
               i + 1, total, average, passed ? "PASS" : "FAIL");
    }
    return 0;
}

LAB A06 · 09

Explain why the program works.

LAB A06 · 10

Build the next skill.

Read the viva answers, try the extra cases, and return to the lab index to mark this exercise as practised.