Matrix transpose using pointer Dry Run in C

Matrix transpose using pointer is an interactive C dry run visualizer from the Pointersarrays programs section. Study the source code, then use the execution controls to follow each step, variable update, highlighted line, and console output.

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Matrix transpose using pointer Program Code

#include <stdio.h>
void transpose(int (*source)[4], int (*dest)[3], int rows, int cols) {
    int j;
    for(i = 0; i < rows; i++) {
        for(j = 0; j < cols; j++) {
            dest[j][i] = source[i][j];
        }
    }
}

void transposePtr(int *src, int *dest, int rows, int cols) {
    int j;
    for(i = 0; i < rows; i++) {
        for(j = 0; j < cols; j++) {
            dest[j * rows + i] = src[i * cols + j];
        }
    }
}

int main() {
    int i, j;
    int matrix[3][4] = {
        {1, 2, 3, 4},
        {5, 6, 7, 8},
        {9, 10, 11, 12}
    };
    int transpose[4][3];
    int rows = 3, cols = 4;
    
    printf("Original matrix (3×4):\n");
    for(i = 0; i < rows; i++) {
        for(j = 0; j < cols; j++) {
            printf("%3d ", matrix[i][j]);
        }
        printf("\n");
    }
    
    // Method 1: Using 2D array pointers
    transpose(matrix, transpose, rows, cols);
    
    printf("\nTranspose (4×3) using row pointer:\n");
    for(i = 0; i < cols; i++) {
        for(j = 0; j < rows; j++) {
            printf("%3d ", transpose[i][j]);
        }
        printf("\n");
    }
    
    // Method 2: Using single pointer
    int transpose2[4][3];
    transposePtr(&matrix[0][0], &transpose2[0][0], rows, cols);
    
    printf("\nTranspose using single pointer:\n");
    for(i = 0; i < cols; i++) {
        for(j = 0; j < rows; j++) {
            printf("%3d ", transpose2[i][j]);
        }
        printf("\n");
    }
    
    return 0;
}

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