C Program to Sort an Array Using Insertion Sort

A class stores five marks in their original order: 40, 35, 45, 35 and 50. The insertion sort task needs a clear rule about order, duplicates and how many elements exist.

Insertion Sort orders a sequence through a defined comparison and preserves its elements, including duplicates.

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 insertion sort uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Order the same marks in the opposite direction using the worked insertion approach.

Insertion Sort: follow the data

Given data
A class stores five marks in their original order: 40, 35, 45, 35 and 50. The insertion sort task needs a clear rule about order, duplicates and how many elements exist.
Rule
Insertion Sort orders a sequence through a defined comparison and preserves its elements, including duplicates.
Observed result
35 35 40 45 50

A guided model of this example. The illustration explains the process; run the complete C program to check its output.

Understand the operation

Keep an ordered prefix and insert each next value by shifting greater prefix elements right. Save the inserted value before making those shifts.

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]);
  for (int i = 1; i < count; i++) {
    int value = marks[i], j = i;
    while (j > 0 && marks[j - 1] > value) { marks[j] = marks[j - 1]; j--; }
    marks[j] = value;
  }
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

Expected output

35 35 40 45 50

Run in C compiler →

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]);
  for (int i = 1; i < count; i++) {
    int value = marks[i], j = i;
    while (j > 0 && marks[j - 1] > value) { marks[j] = marks[j - 1]; j--; }
    marks[j] = value;
  }
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

The second value 35 shifts 40 right; the later duplicate 35 shifts 45 and 40 but not the earlier equal 35.

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 second value 35 shifts 40 right; the later duplicate 35 shifts 45 and 40 but not the earlier equal 35.
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.
3
Check the result
35 35 40 45 50
```
35 35 40 45 50
```

A guided model of this example. The illustration explains the process; run the complete C program to check its output.

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]);
  for (int i = 1; i < count; i++) {
    int value = marks[i], j = i;
    while (j > 0 && marks[j - 1] > value) { marks[j] = marks[j - 1]; j--; }
    marks[j] = value;
  }
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

Order the same marks in the opposite direction using the worked insertion approach.

Worked answer

The worked answer changes the comparison direction while keeping the valid indexes and element count. Equal marks remain present; sorting is not duplicate removal.

C17 · complete program

#include <stdio.h>
int main(void) {
  int marks[] = {40, 35, 45, 35, 50};
  int count = (int)(sizeof marks / sizeof marks[0]);
  for (int i = 1; i < count; i++) {
    int value = marks[i], j = i;
    while (j > 0 && marks[j - 1] < value) { marks[j] = marks[j - 1]; j--; }
    marks[j] = value;
  }
  for (int i = 0; i < count; i++) printf("%d%s", marks[i], i == count - 1 ? "\n" : " ");
  return 0;
}

Expected output

50 45 40 35 35

Run in C compiler →

Recognize the important boundaries

Worst-case moves are quadratic; a mostly ordered list often needs fewer shifts. Strict > shifting preserves the order of equal records.

The sample marks and their sums fit int. General input requires both valid array bounds and an arithmetic-range contract.

Choose the answer, then compare the reasoning for every option.

Use the idea independently

Insertion Sort orders a sequence through a defined comparison and preserves its elements, including duplicates.

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.

Next lesson: C Program to Sort an Array using Merge Sort →

C Program to Sort an Array Using Insertion Sort