My algorithm for transposing matrix only works on certain dimensions-CodePudding

Basically, I have written some code to transpose a matrix and for some inputs it works, for some it does not, i have no clue where did i make a mistake.

This is the code

#include <stdio.h>

int main() {
  int WIDTH, HEIGHT, i, j;
  int matrix[100][100];
  do {
    printf("Enter height and width: ");
    scanf("%d %d", &HEIGHT, &WIDTH);
    if ((WIDTH > 100 || HEIGHT > 100) || (WIDTH < 1) || (HEIGHT < 1))
      printf("Dimensions incorrect!\n");
    else
      break;
  } while (1);

  int vel = WIDTH * HEIGHT;

  printf("Enter elements: ");
  for (i = 0; i < HEIGHT; i  )
    for (j = 0; j < WIDTH; j  )
      scanf("%d", &matrix[i][j]);

  int temp;
  for (i = 0; i < HEIGHT; i  ) {
    for (j = i; j < WIDTH; j  ) {
      temp = matrix[i][j];
      matrix[i][j] = matrix[j][i];
      matrix[j][i] = temp;
    }
  }

  printf("Transposed matrix: \n");
  for (i = 0; i < WIDTH; i  ) {
    for (j = 0; j < HEIGHT; j  )
      printf("]", matrix[i][j]);
    printf("\n");
  }

  return 0;
}

My output for dimensions 3 2, elements {1,2,3,4,5,6}: (example 1)

1    3    0
2    4    0

Wanted output: (example 1)

1    3    5
2    4    6

My output for dimensions 5 1, elements {1,2,3,4,5}: (example 2)

1    0    0    0    0

Wanted output: (example 2)

1    2    3    4     5

CodePudding user response：

Consider the 5,1 case, so:

  for (i = 0; i < 5; i  ) {
    for (j = i; j < 1; j  ) { // for j>=1 the condition will always fail
      temp = matrix[i][j];
      matrix[i][j] = matrix[j][i];
      matrix[j][i] = temp;
    }
  }

The body of the inner loop will only run for [0][0].

That loop will only work if HEIGHT<=WIDTH. If HEIGHT>=WIDTH, you should instead do:

  for (i = 0; i < HEIGHT; i  ) {
    for (j = 0; j < i; j  ) {
      temp = matrix[i][j];
      matrix[i][j] = matrix[j][i];
      matrix[j][i] = temp;
    }
  }

In other words, if the matrix is asymmetric you want to base the number of swaps on its longer side.

If HEIGHT==WIDTH, you can do either one.