Issue with returning a 2 dimensional array

Question

I was asked to write a program that gets a two dimensional array (a matrix), number of columns,and number of rows, and the program will return the transpose matrix (without using a [][], meaning only using pointer arithmetics)

The program I wrote, does indeed transpose the matrix, it is no problem. My problem is understanding how to return. here's my code:

int** transpose_matrix(matrix mat1,int number_of_rows,int number_of_columns)
{
    matrix mat2;
    int row_index,column_index;
    for(row_index=0;row_index<number_of_rows;row_index++)
    {
        for(column_index=0;column_index<number_of_columns;column_index++)
            **(mat2+(column_index*number_of_rows)+row_index)=**(mat1+(row_index*number_of_columns)+column_index);
    }
    // at this point, mat2 is exactly the transpose of mat1
    return mat2;
}

now here's my problem: I can't return a matrix, closest thing I can do is return the address of the first value of the matrix, but even if i do that, all the rest of the matrix will be unusable as soon as i exit transpose_matrix function back into void main...How can I return mat2?

@nhahtdh malloc does not allocate space on the stack. – Carey Gregory Mar 19 '13 at 20:44 — Carey Gregory, Mar 19 '13 at 20:44
@nhahtdh It's the heap that is dynamic. – squiguy Mar 19 '13 at 20:50 — squiguy, Mar 19 '13 at 20:50
Think of heap, but written stack. #_# – nhahtdh Mar 19 '13 at 22:30 — nhahtdh, Mar 19 '13 at 22:30

score 0 · Accepted Answer · 2013-03-19T21:09:08.063

0

One, a two-dimensional array is not a double pointer.

Two, dynamic allocation. If matrix is a two-dimensional array type, then write something like this:

typedef int matrix[ROWS][COLUMNS];
typedef int (*matrix_ptr)[COLUMNS];

matrix_ptr transpose_matrix(matrix m, int rows, int cols)
{
    matrix_ptr transposed = malloc(sizeof(*transposed) * rows);
    // transpose, then
    return transposed;
}

edited Mar 19 '13 at 21:09

answered Mar 19 '13 at 20:40

@qPCR4vir Why are you shouting? – Mar 19 '13 at 21:03
Anyway, I want to make +1 you, but I want to see the "// transpose, then " part first. – qPCR4vir Mar 19 '13 at 21:13
@qPCR4vir Sorry but that's not relevant. OP has already written the code for transposition, he was merely asking for a way to return an entire matrix. – Mar 19 '13 at 21:17
@qPCR4vir Why wouldn't it? – Mar 19 '13 at 21:20
Maybe Im wrong, but '**(transposed+i)' is '*((int**)transposed+i*COLUMNS)' ?? – qPCR4vir Mar 19 '13 at 21:26
@qPCR4vir What you wrote is a syntax error. `transposed` is a [pointer-to-array](http://c-faq.com/aryptr/ptrtoarray.html). – Mar 19 '13 at 21:28
Yes that ias exactly the problem. The OP code is for pointer to int, and not for a pointer to an array of COLUMNS int (to an array of ROWS int could be better?? it is stransposed). – qPCR4vir Mar 19 '13 at 21:34
Guys I'm really new at all of this, can you explain to me what the argument is about? maybe I can learn something. And also, explain the code you wrote. Assume I'm stupid. (and the code for my transpose does work, the problem is with the return) – Oria Gruber Mar 19 '13 at 21:35
@qPCR4vir OP's code assumes that the matrix is represented as a 2D array. Unless you want to manually spell out ugly indexing like `ptr[rows * column + row]`, you can use a pointer-to-array and let the compiler do the indexing for you safely. Whether the pointer is to `int[ROWS]` or to `int[COLUMNS]` entirely depends on the code OP has so far. – Mar 19 '13 at 21:39

qPCR4vir · Answer 2 · 2013-03-19T22:45:08.410

OK: Here you have 3 thing:

You cannot return a pointer to a local variable (it will be garbage after return and the stack (memory) where it was is reused).
Array decay to pointer to the first element when passed.
Pointer arithmetic: p+1 increment the adress in p by sizeof(*p), so p point to the next element, not to the next byte.

The simple fix for your code (this works for any matrix size) :

    int* transpose_matrix(int *mat1,int number_of_rows,int number_of_columns)
    {
        int *mat2=malloc(number_of_rows*number_of_columns*sizeof(int));
        int row_index,column_index;
        for(row_index=0;row_index<number_of_rows;row_index++)
        {
            for(column_index=0;column_index<number_of_columns;column_index++)
                mat2[column_index*number_of_rows+row_index]=mat1[row_index*number_of_columns+column_index];
        }
        // at this point, mat2 is exactly the transpose of mat1
        return mat2;
    }

...
    print(m,r,c); // I hope you have a print()
    int *t=transpose_matrix(m,r,c);
    print (t,c,r);
...
    // use t[max: c-1][max: r-1]
    free(t);

If we have only fixed size matrix (well with C99 we can use varable length array too!).

typedef int  Matrix[ROWS][COLUMNS];
typedef int TMatrix[COLUMNS][ROWS];
typedef int (*pMatrix)[COLUMNS]; 
typedef int (*pTMatrix)[ROWS];  

pTMatrix transpose_matrix(Matrix m , int rows, int cols)
{
    pTMatrix t = malloc(sizeof(*t)*cols);
    for (int r=0; r<rows ; ++r)
    for (int c=0; r<cols ; ++r)
      t[c][r]=m[r][c];
    return t;
}

Well, if rows and cols are fixed you dont need to pass it.... hmmm...

Okay, this seems to work, but I have another problem now, I need to print the transpose matrix, but you have declared m to be a pointer to an integer, so you've turned the 2d array into a 1d array, how do I print the transpose matrix? — Oria Gruber, Mar 19 '13 at 21:56

Issue with returning a 2 dimensional array

2 Answers2