Transposing matrix

**2112** · 02-08-2008

Hi all,
I'm learning C by myself and, as exercise, I was trying to make a program able to transpose a matrix[a][b] to matrix[b][a]. I can't really understand what's wrong and sadly this exercise has no solution posted on the net that i could check. It compiles fine on Pelles C, but the displayed result is a bit strange.

Code:

#include <stdio.h>

void displayMatrix (int x, int y, int M3[x][y])
{
	int a = 0, b = 0;

	for ( a = 0; a <= x; ++a) {
		for ( b = 0; b <= y; ++b)
			printf("&#37;5i", M3[a][b]);
	printf("\n");
	}
}

void transposeMatrix (int x, int y, int M1[x][y], int w, int z,int M2[w][z])
{
	int a = 0, b = 0;

	for ( a = 0; a <= x; ++a) {
		for ( b = 0; b <= y; ++b)
			M2[b][a] = M1[a][b];
	}
}


int main (void)
{
	void transposeMatrix (int x, int y, int M1[x][y], int w, int z,int M2[w][z]);
	void displayMatrix (int x, int y, int M3[x][y]);
	
	int Matrix2[5][4];

	int Matrix1[4][5] = 
	{ 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }
	};

	displayMatrix(4, 5, Matrix1);  //test display function -- seems to work yahoo..

	printf("\n\n-----------------------------------\n\nTransposed matrix:\n\n");  // some space

	transposeMatrix(4, 5, Matrix1, 5, 4, Matrix2);

	displayMatrix(5, 4, Matrix2);

	return 0;
}

I used void as return value for transposeMatrix cause I guess matrix behave like arrays, the matrix gets directly modified by the function? no need to return in theory? (even it might be good practice) Sorry for the maybe silly question but i just arrived to multi dimensional arrays and i can't quite figure it out myself.
Thanks in advance.

**matsp** · 02-08-2008

Yes, you don't have to (and it's probably meaningless to) return anything from transposeMatrix.

However, I'd like to point out: The C language doesn't normally allow variable size arrays, so

Code:

void transposeMatrix (int x, int y, int M1[x][y], int w, int z,int M2[w][z])

wouldn't be valid in all compilers - not sure if it is in ANY compiler, really, since even newer compilers expect y and z to be constants [x and w are not even needed in the parameter declaration].

--
Mats

**2112** · 02-08-2008

Thanks for the answer Mats.
Although if i use normal declaration not only i get funny results but i get "application error" and this is a first for me eheheh. I dont know there's still something wrong I cannot see. Still looking into it. I wonder if that a = 0 must be a = 1 but it doesn't sem to work either...

Code:

#include <stdio.h>

void displayMatrix (int M3[5][4])
{
	int a = 0, b = 0;

	for ( a = 0; a <= 5; ++a) {
		for ( b = 0; b <= 4; ++b)
			printf("%5i", M3[a][b]);
	printf("\n");
	}
}

void transposeMatrix (int M1[4][5], int M2[5][4])
{
	int a = 0, b = 0;

	for ( a = 0; a <= 4; ++a) {
		for ( b = 0; b <= 5; ++b)
			M2[b][a] = M1[a][b];
	}
}


int main (void)
{
	void transposeMatrix (int M1[4][5], int M2[5][4]);
	void displayMatrix (int M3[5][4]);
	
	int Matrix2[5][4];

	int Matrix1[4][5] = 
	{ 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }
	};

	// displayMatrix(Matrix1);  //test display function -- seems to work yahoo..

	printf("\n\n-----------------------------------\n\nTransposed matrix:\n\n");  // some space

	transposeMatrix(Matrix1, Matrix2);

	displayMatrix(Matrix2);

	return 0;
}

**matsp** · 02-08-2008

Code:

	for ( a = 0; a <= 4; ++a) {
		for ( b = 0; b <= 5; ++b)
			M2[b][a] = M1[a][b];
	}

You are copying one cell too many in each direction, since you use a <= 4 and b <= 5. Cells are 0..3 when you have array[4], and 0..4 if you use array[5] - you are setting cells 0..4 and 0..5 respectively.

--
Mats

**2112** · 02-08-2008

Ah damn how could I forget that, that's a basic of arrays...whoops it's that i studied arrays and functions all in a few days and I have yet to get used to it. Thanks!

**2112** · 02-08-2008

On a side note, with that corrected, it seems to work also using variable length arrays:

Code:

#include <stdio.h>

void displayMatrix (int x, int y, int M3[x][y])
{
	int a = 0, b = 0;

	for ( a = 0; a <= x; ++a) {
		for ( b = 0; b <= y; ++b)
			printf("&#37;5i", M3[a][b]);
	printf("\n");
	}
}

void transposeMatrix (int x, int y, int M1[x][y], int w, int z,int M2[w][z])
{
	int a = 0, b = 0;

	for ( a = 0; a <= x; ++a) {
		for ( b = 0; b <= y; ++b)
			M2[b][a] = M1[a][b];
	}
}


int main (void)
{
	void transposeMatrix (int x, int y, int M1[x][y], int w, int z,int M2[w][z]);
	void displayMatrix (int x, int y, int M3[x][y]);
	
	int Matrix2[5][4];

	int Matrix1[4][5] = 
	{ 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }, 
		{ 1, 2 , 3, 4, 5 }
	};

	displayMatrix(4, 5, Matrix1);  //test display function -- seems to work yahoo..

	printf("\n\n-----------------------------------\n\nTransposed matrix:\n\n");  // some space

	transposeMatrix(4, 5, Matrix1, 5, 4, Matrix2);

	displayMatrix(5, 4, Matrix2);

	return 0;
}

compiles on Pelles C but not on GCC stable, maybe using the candidate package dont know

**matsp** · 02-08-2008

Sure, if you want to lock yourself into gcc, then fine.

--
Mats

**2112** · 02-08-2008

locked in GCC? not really

In GCC i get "internal compiler error: in put_pending_sizes, at stor-layout.c:140"
Well that surely doesn't make me wanna swap from Pelles to GCC...although in newer version it's probably fixed. The stable is pretty old by now.

**matsp** · 02-08-2008

What I mean is that you are locked into some particular compiler versions that allow this construct. Many compilers won't.

--
Mats

**2112** · 02-08-2008

ahhh yea but it's fine for now, i'm only learning so i need also to test out features like variable size arrays,
When i will make my first real project, then ill take that into account for sure

**DQNOK** · 02-08-2008

I'm stunned that the code works at all. If you understand how multidimensional arrays actually work (and that can be a pretty confusing subject), then you understand why, when passing an array to a function, only the first index can be left blank; all the others must be populated with constants.

Multidimensional arrays are actually stored in contiguous memory as though it was a single-dimensional array. The compiler must know the size of all the last dimensions in order to locate the element you are referencing.

I long ago quit using multidimensional arrays, and instead went to macros that emulate the same behavior, but are more "dynamic".

Code:

void matrixprocess( double* matrix, size_t numrows, size_t numcols )
{
   #define M(i,j) (matrix[(i)*numcols + (j)])
/*Note that i and j are zero-based indices, not like in Fortran. */
...
   #undef M
}

This code works regardless of the number of columns and rows.

DQNOK

**tabstop** · 02-08-2008

In C, you can only leave the first index blank; all the last dimensions must be specified. Your macro at least gets it right.

**matsp** · 02-08-2008

Actually, it's the FIRST dimension that can be left out (and is not checked by the compiler anyways). All other dimensions need to be passed.

Apparently the compiler used in this case can cope with variable size arrays, and that should work both with passed arguments and local variables. But this is, as far as I know, an extension on top of C99.

--
Mats

**laserlight** · 02-08-2008

Apparently the compiler used in this case can cope with variable size arrays, and that should work both with passed arguments and local variables. But this is, as far as I know, an extension on top of C99.

I had the impression that variable length arrays were a feature, not extension, of C99.