D13 Answer

#define NUMCOLS 5

int transpose( int x[NUMCOLS][NUMCOLS] )
{
  int r,  c, temp ;

  for ( r=0; r<NUMCOLS; r++ )
    for ( c=0; c<r; c++ )
    {
      temp    = x[r][c];
      x[r][c] = x[c][r];
      x[c][r] = temp;
    }
}

Remarks: A good number of you probably made this mistake:

#define NUMCOLS 5

int transpose( int x[NUMCOLS][NUMCOLS] )
{
  int r,  c, temp ;

  for ( r=0; r<NUMCOLS; r++ )
    for ( c=0; c< NUMCOLS; c++ )
    {
      temp    = x[r][c];
      x[r][c] = x[c][r];
      x[c][r] = temp;
    }
}

and discovered that the 2D array did not change. Of course, this is because when c is to the right of the diagonal, the function swaps two elements that have already been swapped. In effect, the fuction transposes the array twice. The transpose of the transpose is the same as the original.

This solution ignores the fact that diagonal elements do not change. When r==c, the body of the loop exchanges and single element (on the diagonal) with itself. This does not hurt anything. You might be tempted to write code that avoids this needless work, and write:

int transpose( int x[][NumColumns], int nrows )
{
  int r,  c, temp ;

  for ( r=0; r<NUMCOLS; r++ )
    for ( c=0; c<r; c++ )
      if ( r!=c )
      {
        temp    = x[r][c];
        x[r][c] = x[c][r];
        x[c][r] = temp;
      }
}

But now the extra work of computing ( r!=c ) is done every time the loop body executes, not just for the relatively rare diagonal elements. This is much more work than the needless work done along the diagonal. But in either case, the extra work is trivial.