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.
