matlab engine giving wrong result when called in a function in Cpp file

matlab engine call inside function: i try to sort an array of numbers & write the result to a file(using dlmwrite). also i display them on the screen. * when i sort { 0, 5, 2, 7, 3, 9, 1, 6, 8, 4 } i get an array of ten zeros(both at std::cout & using dlmwrite)*

my code for the function:

int call_matlab_processing(double double_array[], int size_double_array)
{
  // its just a sorting algo, sorting being implemented by matlab
  double * sorted;

  Engine *ep;
  mxArray *T = NULL, *result = NULL;
  if (!(ep = engOpen(""))) {
    fprintf(stderr, "\nCan't start MATLAB engine\n");
    return 1;
  }
  T = mxCreateDoubleMatrix(1,size_double_array, mxREAL);
  memcpy((void *)mxGetPr(T), (void *)double_array, sizeof(double_array));

  engPutVariable(ep, "T", T);
  engEvalString(ep, "[D] = sort(T,'descend')");
  engEvalString(ep, "dlmwrite('myFile.txt',D)"); * * *// writes 0 0 0 ... 0 to file***
  result = engGetVariable(ep,"D");

  sorted=(double *)mxGetData(result);
   * *// i try to print the contents of the sorted array, but it gives 0 0 0 ...0**
  for(int i = 0; i < size_double_array;i++)
  {
    std::cout<< "double_array at "<< i<< "="<< *sorted<< std::endl;
    sorted++;
  }

  mxDestroyArray(T);
  mxDestroyArray(result);

  engEvalString(ep, "close;");
  engClose(ep);

  return 0;
}

Answer 1

[James Tursa]

These lines:

int call_matlab_processing(double double_array[], int size_double_array,double * sorted, double * indices_cpp)
{
    :
  memcpy((void *)mxGetPr(T), (void *)double_array, sizeof(double_array));

The first argument of the function call_matlab_processing is the variable double_array. It turns out that this variable is of type "pointer to double" (i.e., double * ). I know it looks like you have declared it as an array, but that syntax is misleading. When used in a function argument, the notation

double variable_name[]

is equivalent to the notation

double *variable_name

In fact, even if you had use an explicit size it would have made no difference to the compiler. I.e., this notation in a function argument

double variable_name[10]

is also equivalent to the notation

double *variable_name

That is, the compiler sees that argument as a pointer, not as an array (you can't pass whole arrays in C/C++ function arguments this way). So downstream in your code when you use sizeof(double_array), it is equivalent to doing sizeof(double * ). The result will be either 4 (on 32-bit) or 8 (on 64-bit). So you are definitely not copying all of the elements in that memcpy call. You need to do this instead:

  memcpy((void *)mxGetPr(T), (void *)double_array, size_double_array*sizeof(double));

Also, FYI you don't really need the (void *) casts since converting pointers to/from void * is something the C/C++ compiler will automatically do for you.

Answer 2

[Titus Edelhofer]

Hi,

first of all, your memcpy copies just 4 or 8 bytes but not the array:

memcpy((void *)mxGetPr(T), (void *)double_array, sizeof(double_array));

You need to replace by

memcpy((void *)mxGetPr(T), (void *)double_array, size_double_array * sizeof(double));

And for the result you would need to do:

double *sorted;
sorted = mxGetPr(result);
for (int i=0; i<size_double_array; i++) {
   std::cout << sorted[i] << endl;
}

Titus

[James Tursa]

He is printing out *sorted and doing sorted++ inside the loop, so he should get the same result as sorted[i].

[Titus Edelhofer]

James, you are of course right...