C Program to Find Average of Array Elements
C Program to Find Average of Array Elements
A class stores five marks in their original order: 40, 35, 45, 35 and 50. The average array mark task needs a clear rule about order, duplicates and how many elements exist.
An arithmetic mean divides the sum of all included values by their count, with floating-point division when a fractional result is needed.
Before this exercise, review array elements and bounds and function parameters. The worked program below applies those ideas to this specific problem.
Explain
Explain why average array mark uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Print the numerator and denominator separately so the mean calculation can be audited.
Average Array Mark: follow the data
Given data
A class stores five marks in their original order: 40, 35, 45, 35 and 50. The average array mark task needs a clear rule about order, duplicates and how many elements exist.
Rule
An arithmetic mean divides the sum of all included values by their count, with floating-point division when a fractional result is needed.
Observed result
Average: 41.0
Understand the operation
Keep sum and count attached to the same population. Convert before division; storing an already-truncated integer quotient in double does not restore its fraction.
The complete example knows its five-element bound. In a reusable function, pass an explicit element count with the array pointer; a pointer alone does not report that bound. This sample has two equal marks, which makes duplicate handling observable.
Read the complete C program
The input values are written into this complete C17 example so the calculation can be reproduced without an unavailable service. Change the named data, rebuild and compare the actual output with your prediction.
C17 · complete program
#include <stdio.h>
int main(void) {
int marks[] = {40, 35, 45, 35, 50};
int count = (int)(sizeof marks / sizeof marks[0]);
int total = 0;
for (int i = 0; i < count; i++) total += marks[i];
printf("Average: %.1f\n", (double)total / count);
return 0;
}
Expected output
Average: 41.0
Follow the values through the code
C17 · code used in this explanation
#include <stdio.h>
int main(void) {
int marks[] = {40, 35, 45, 35, 50};
int count = (int)(sizeof marks / sizeof marks[0]);
int total = 0;
for (int i = 0; i < count; i++) total += marks[i];
printf("Average: %.1f\n", (double)total / count);
return 0;
}
The running total is 40, 75, 120, 155 and 205. Dividing 205 by five gives 41.0.
Compare the resulting output with the original sequence. A changed order and a changed set of values are different effects; decide which your operation promises.
Follow the worked calculation
Start with the sample
The running total is 40, 75, 120, 155 and 205. Dividing 205 by five gives 41.0.
Apply the rule
Compare the resulting output with the original sequence. A changed order and a changed set of values are different effects; decide which your operation promises.
Check the result
Average: 41.0
Average: 41.0
Try a small variation
C17 · code used in this explanation
#include <stdio.h>
int main(void) {
int marks[] = {40, 35, 45, 35, 50};
int count = (int)(sizeof marks / sizeof marks[0]);
int total = 0;
for (int i = 0; i < count; i++) total += marks[i];
printf("Average: %.1f\n", (double)total / count);
return 0;
}
Print the numerator and denominator separately so the mean calculation can be audited.
Worked answer
The complete answer keeps the five valid input elements and applies the stated alternative. Read the output against the original array rather than assume that every loop should sort, remove or overwrite values.
C17 · complete program
#include <stdio.h>
int main(void) {
int marks[] = {40, 35, 45, 35, 50};
int count = (int)(sizeof marks / sizeof marks[0]);
int total = 0;
for (int i = 0; i < count; i++) total += marks[i];
printf("Total: %d; count: %d\n", total, count);
return 0;
}
Expected output
Total: 205; count: 5
Recognize the important boundaries
Reject an empty population before division. A larger dataset can overflow a narrow sum even if every individual mark is valid.
The sample marks and their sums fit int. General input requires both valid array bounds and an arithmetic-range contract.
Check your understanding
Choose the answer, then compare the reasoning for every option.
Use the idea independently
An arithmetic mean divides the sum of all included values by their count, with floating-point division when a fractional result is needed.
Continue with array indexes and bounds.
Continue with pointer and count contracts.
Language reference: WG14 C11 committee draft. These examples use the C17-compatible subset of the explained language features.
Reaching the recap and completing the quizzes are separate milestones.