C Program to Find the Transpose of a Matrix
C Program to Find the Transpose of a Matrix
A small school table has two rows and two columns. The transpose a matrix operation should preserve what each row and column means, rather than treating the table as four unrelated values.
Transposition exchanges row and column positions: destination[col][row] receives source[row][col].
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 transpose a matrix uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Transpose the result again and verify that the original arrangement returns.
Transpose a Matrix: follow the data
Given data
A small school table has two rows and two columns. The transpose a matrix operation should preserve what each row and column means, rather than treating the table as four unrelated values.
Rule
Transposition exchanges row and column positions: destination[col][row] receives source[row][col].
Observed result
1 3 2 4
Understand the operation
A table whose rows describe classes and columns describe days can be viewed with those roles exchanged. For a rectangular table, the output dimensions are reversed.
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}};
int transposed[2][2];
for (int row = 0; row < 2; row++) for (int col = 0; col < 2; col++) transposed[col][row] = matrix[row][col];
for (int row = 0; row < 2; row++) { for (int col = 0; col < 2; col++) printf("%d%s", transposed[row][col], col == 1 ? "\n" : " "); }
return 0;
}
Expected output
1 3
2 4
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}};
int transposed[2][2];
for (int row = 0; row < 2; row++) for (int col = 0; col < 2; col++) transposed[col][row] = matrix[row][col];
for (int row = 0; row < 2; row++) { for (int col = 0; col < 2; col++) printf("%d%s", transposed[row][col], col == 1 ? "\n" : " "); }
return 0;
}
The source value two at row zero, column one moves to row one, column zero.
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 source value two at row zero, column one moves to row one, column zero.
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
1 3 2 4
1 3 2 4
Try a small variation
C17 · code used in this explanation
#include <stdio.h>
int main(void) {
int matrix[2][2] = {{1, 2}, {3, 4}};
int transposed[2][2];
for (int row = 0; row < 2; row++) for (int col = 0; col < 2; col++) transposed[col][row] = matrix[row][col];
for (int row = 0; row < 2; row++) { for (int col = 0; col < 2; col++) printf("%d%s", transposed[row][col], col == 1 ? "\n" : " "); }
return 0;
}
Transpose the result again and verify that the original arrangement returns.
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}};
int transposed[2][2];
for (int row = 0; row < 2; row++) for (int col = 0; col < 2; col++) transposed[col][row] = matrix[row][col];
for (int row = 0; row < 2; row++) for (int col = 0; col < 2; col++) matrix[col][row] = transposed[row][col];
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
Writing directly into the source without a correct in-place strategy can overwrite values that still need to be read.
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
Transposition exchanges row and column positions: destination[col][row] receives source[row][col].
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.