From help-request at octave dot org Wed Jun  2 08:39:26 2004
Subject: Re: A question about programming using dynamically linked function
From: Paul Thomas <paulthomas2 at wanadoo dot fr>
To: jiaguo at uiuc dot edu
Cc: help at octave dot org
Date: Wed, 02 Jun 2004 15:38:33 +0200

Find below a slightly different approach to David's.  Like David the 
Array<T> function fortran_vec() is used to point to the first member in 
the complexMatrix.  However, my example uses the C++ standard library 
complex, rather than the octave Complex.  I have commented out a 
transpose that operates directly on the complex matrices and replaced it 
in the next line with the same operation using complex pointers.

Best regards

Paul T

#include <octave/oct.h>
#include <octave/quit.h>
#include <complex>
typedef std::complex<double> dComplex;
DEFUN_DLD (complexdemo, args, ,
"Demonstrate fortran_vec for complex \n\n\
transpose_of_x=complexdemo(x) \n") {
  if ( args.length() !=  1 ) {
    error("this version of matrixdemo omly takes one argument");
    return octave_value( 0 );
  }
  ComplexMatrix xin1( args(0).complex_matrix_value() );
  int ir = xin1.rows();
  int ic = xin1.cols();
  ComplexMatrix xout1(ic,ir);
  dComplex *pin1 = xin1.fortran_vec();
  dComplex *pout1 = xout1.fortran_vec();
  for (int ir1 = 0 ; ir1 < ir ; ir1++)
    for (int ic1 = 0 ; ic1 < ic ; ic1++) {
//      xout1(ic1,ir1) = xin1(ir1,ic1);                //transpose using 
Matrices
       pout1[ic1 + ir1*ic] = pin1[ir1 + ic1*ir];       //transpose using 
pointers
    }
  return octave_value( xout1 );
}  



jiaguo at uiuc dot edu wrote:

>Hi
>
>I am writing dynamically linked functions. In my function, I
>need to copy the data from a complex matrix (in octave_value
>type) to a double buffer. I was wondering if there is a way to
>find a pointer to the real data in the complex matrix (in
>octave_value type), so that I can use memcpy to copy it to my
>own buffer. The similar function can be found in MATLAB which
>is mxGetPr. Does Octave have similar functions like mxGetPr?
>
>Thanks a lot!
>
>Jia
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