C Program to Swap First and Last Matrix Elements
C Program to Swap First and Last Matrix Elements
A small school table has two rows and two columns. The swap row endpoints operation should preserve what each row and column means, rather than treating the table as four unrelated values.
Swap the first and last entries within each row, preserving row membership and using temporary storage.
Before this exercise, review array indexes and bounds and nested loops. The worked program below applies those ideas to this specific problem.
Explain
Explain why swap row endpoints uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Apply the same swap twice and verify that each row returns to its original arrangement.
Swap Row Endpoints: follow the data
Given data
A small school table has two rows and two columns. The swap row endpoints operation should preserve what each row and column means, rather than treating the table as four unrelated values.
Rule
Swap the first and last entries within each row, preserving row membership and using temporary storage.
Observed result
2 1 4 3
Understand the operation
For a two-column row, endpoint swapping also reverses the whole row. For wider rows, the interior columns are left unchanged.
The example fixes both dimensions at two so every index can be checked directly. A general matrix interface must carry dimensions, require compatible shapes for its operation and validate storage for each referenced element.
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 matrix[2][2] = {{1, 2}, {3, 4}};
for (int row = 0; row < 2; row++) { int saved = matrix[row][0]; matrix[row][0] = matrix[row][1]; matrix[row][1] = saved; }
for (int row = 0; row < 2; row++) {
for (int col = 0; col < 2; col++) printf("%d%s", matrix[row][col], col == 1 ? "\n" : " ");
}
return 0;
}
Expected output
2 1
4 3
Follow the values through the code
C17 · code used in this explanation
#include <stdio.h>
int main(void) {
int matrix[2][2] = {{1, 2}, {3, 4}};
for (int row = 0; row < 2; row++) { int saved = matrix[row][0]; matrix[row][0] = matrix[row][1]; matrix[row][1] = saved; }
for (int row = 0; row < 2; row++) {
for (int col = 0; col < 2; col++) printf("%d%s", matrix[row][col], col == 1 ? "\n" : " ");
}
return 0;
}
The first row exchanges one and two; the second exchanges three and four.
Read the resulting rows in order; moving a value to another row or column is part of the operation’s meaning, not merely an output-format choice.
Follow the worked calculation
Start with the sample
The first row exchanges one and two; the second exchanges three and four.
Apply the rule
Read the resulting rows in order; moving a value to another row or column is part of the operation’s meaning, not merely an output-format choice.
Check the result
2 1 4 3
2 1 4 3
Try a small variation
C17 · code used in this explanation
#include <stdio.h>
int main(void) {
int matrix[2][2] = {{1, 2}, {3, 4}};
for (int row = 0; row < 2; row++) { int saved = matrix[row][0]; matrix[row][0] = matrix[row][1]; matrix[row][1] = saved; }
for (int row = 0; row < 2; row++) {
for (int col = 0; col < 2; col++) printf("%d%s", matrix[row][col], col == 1 ? "\n" : " ");
}
return 0;
}
Apply the same swap twice and verify that each row returns to its original arrangement.
Worked answer
The complete worked answer uses the same valid two-by-two input. Trace its index rule independently; compare each printed result with the source element or accumulation that produced it.
C17 · complete program
#include <stdio.h>
int main(void) {
int matrix[2][2] = {{1, 2}, {3, 4}};
for (int row = 0; row < 2; row++) { int saved = matrix[row][0]; matrix[row][0] = matrix[row][1]; matrix[row][1] = saved; }
for (int row = 0; row < 2; row++) { int saved = matrix[row][0]; matrix[row][0] = matrix[row][1]; matrix[row][1] = saved; }
for (int row = 0; row < 2; row++) {
for (int col = 0; col < 2; col++) printf("%d%s", matrix[row][col], col == 1 ? "\n" : " ");
}
return 0;
}
Expected output
1 2
3 4
Recognize the important boundaries
A zero-column row has no endpoints. One-column endpoint swapping is a no-operation on the same position.
Larger or untrusted dimensions require checked storage-size calculations. Integer products and accumulated sums also need a representable-range contract.
Check your understanding
Choose the answer, then compare the reasoning for every option.
Use the idea independently
Swap the first and last entries within each row, preserving row membership and using temporary storage.
Continue with array boundaries.
Continue with function bounds 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.