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finding matrix transpose - why doesn't it work when passing pointer argument?

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finding matrix transpose - why doesn't it work when passing pointer argument?

 
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  #1
Feb 7th, 2008
Hi

I'm adapting some code I've written using 2d arrays (to represent matrices) to handle large arrays such that double matrix[][] goes to double **matrix and then I'm using malloc.

It seems to work fine for part of my program up to where I have to find the matrix transpose at which point it does something I don't understand. I've taken that bit of code out and run it by itself (included below), get the same problem and still can't work out why. Can anyone help??
Thanks
jbd
c Syntax (Toggle Plain Text) 
  1. int find_transpose(int n, double **a, double **b)
  2. {
  3.  int i,j;
  4.  for (i=0; i<n; i++)
  5.   {
  6.   for (j=0; j<n; j++)
  7.   b[i][j] = a[i][j];
  8.   }
  9.  for (i=0; i<n; i++)
  10.   {
  11.   for (j=0; j<n; j++)
  12.   b[i][j] = a[j][i];
  13.   }
  14.  
  15. }
  16.  
  17.  
  18. int main (void)
  19. {
  20.  
  21.  int i, j, p=3;
  22.  double **in, **out;
  23.  
  24.  in = malloc(p * sizeof(int *));
  25.  out = malloc(p * sizeof(int *));
  26.  
  27.  for (i=0;  i<p; i++){
  28.  in[i]= malloc(p * sizeof(int *));
  29.  out[i]= malloc(p * sizeof(int *));}
  30.  
  31.   for (i=0;  i<p; i++)
  32.   {
  33.    for (j=0;  j<p; j++)
  34.    {in[i][j]= 10./(i+1);
  35.    printf("in[%i][%i] = %f\n", i, j, in[i][j]);
  36.    }
  37.  }
  38.  
  39.  find_transpose(p, in, out);
  40.  
  41.  for (i=0;  i<p; i++)
  42.   {
  43.    for (j=0;  j<p; j++)
  44.    printf("in[%i][%i] = %f\n", i, j, in[i][j]);
  45.   }
  46.  
  47.  for (i=0;  i<p; i++)
  48.   {
  49.    for (j=0;  j<p; j++)
  50.    printf("out[%i][%i] = %f\n", i, j, out[i][j]);
  51.   }
  52.  
  53. return
Last edited by Narue; Feb 7th, 2008 at 12:28 pm. Reason: Added code tags, do it yourself next time, please.
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Re: finding matrix transpose - why doesn't it work when passing pointer argument?

 
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  #2
Feb 7th, 2008
What does not work? Does it gives you compile errors? Does it crashes?
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Re: finding matrix transpose - why doesn't it work when passing pointer argument?

 
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  #3
Feb 7th, 2008
I corrected minor syntactical errors in your code. I did not check the logic. Check following code.

C Syntax (Toggle Plain Text) 
  1. #include<stdio.h>
  2. #include<stdlib.h>
  3.  
  4. int find_transpose(int n, double **a, double **b)
  5. {
  6. 	int i,j;
  7. 	for (i=0; i<n; i++)
  8. 	{
  9. 		for (j=0; j<n; j++)
  10. 			b[i][j] = a[i][j];
  11. 	}
  12. 	for (i=0; i<n; i++)
  13. 	{
  14. 		for (j=0; j<n; j++)
  15. 			b[i][j] = a[j][i];
  16. 	}
  17. 	return 0;
  18. }
  19.  
  20.  
  21. int main (void)
  22. {
  23.  
  24. 	int i, j, p=3;
  25. 	double **in, **out;
  26.  
  27. 	in = (double **)malloc(p * sizeof(int *));
  28. 	out = (double **)malloc(p * sizeof(int *));
  29.  
  30. 	for (i=0; i<p; i++){
  31. 		in[i]= (double *)malloc(p * sizeof(int *));
  32. 		out[i]= (double *)malloc(p * sizeof(int *));}
  33.  
  34. 	for (i=0; i<p; i++)
  35. 	{
  36. 		for (j=0; j<p; j++)
  37. 		{in[i][j]= 10./(i+1);
  38. 		printf("in[%i][%i] = %f\n", i, j, in[i][j]);
  39. 		}
  40. 	}
  41.  
  42. 	find_transpose(p, in, out);
  43.  
  44. 	for (i=0; i<p; i++)
  45. 	{
  46. 		for (j=0; j<p; j++)
  47. 			printf("in[%i][%i] = %f\n", i, j, in[i][j]);
  48. 	}
  49.  
  50. 	for (i=0; i<p; i++)
  51. 	{
  52. 		for (j=0; j<p; j++)
  53. 			printf("out[%i][%i] = %f\n", i, j, out[i][j]);
  54. 	}
  55.  
  56. 	return 0;
  57. }
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Re: finding matrix transpose - why doesn't it work when passing pointer argument?

 
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  #4
Feb 7th, 2008
Still getting same problem... which is that the input matrix is getting modified when the function is called, here's an example for a test 3 by 3:
The input is:in[0][0] = 10.000000
in[0][1] = 10.000000
in[0][2] = 10.000000
in[1][0] = 5.000000
in[1][1] = 5.000000
in[1][2] = 5.000000
in[2][0] = 3.333333
in[2][1] = 3.333333
in[2][2] = 3.333333

but after being passed to find_transpose is comes out as:in[0][0] = 10.000000
in[0][1] = 10.000000
in[0][2] = 10.000000
in[1][0] = 10.000000
in[1][1] = 5.000000
in[1][2] = 10.000000
in[2][0] = 3.333333
in[2][1] = 3.333333
in[2][2] = 10.000000

and the actual transpose output is:
out[0][0] = 10.000000
out[0][1] = 10.000000
out[0][2] = 10.000000
out[1][0] = 10.000000
out[1][1] = 5.000000
out[1][2] = 3.333333
out[2][0] = 10.000000
out[2][1] = 10.000000
out[2][2] = 10.000000

I really don't understand why!?

Thanks
jbd
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Re: finding matrix transpose - why doesn't it work when passing pointer argument?

 
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  #5
Feb 7th, 2008
Watch your types.

Going off of dubydapreek's revision:
  • Lines 7..11: Useless. Get rid of them. (You don't need to copy b to a before you copy b to a.)
  • Lines 27..28: You should be allocating an array of (double *).
  • Lines 31..32: You should be allocating an array of (double).

As a rule, when using malloc(), it should look something like:
type *foo = (type *)malloc( n *sizeof( type ) );
where "type" is whatever type you need it to be. In your case, "type" is first (double*) then it is (double).

Hope this helps.
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