c-fortran互操作性-带有指针的派生类型

Question

I have long fortran code that has to be made usable from python. I decided to do Fortran->C->Python interface.

I got a problem: I have derived types within Fortran modules that contain

double precision, allocatable  

type as members.

When trying to compile with ifort I get (with gfortran something similar):

Each component of a derived type with the BIND attribute shall be a nonpointer,         
nonallocatable data component with interoperable type and type parameters

This is actually with agreement with Intel compiler documentation and Fortran 2003 standard: point 15.2.5.

Is there any way to access Fortran "type" with allocatable or pointer inside?

Answer 1

Not really. Even the brand new TS 29113 does not offer a solution. It has the TYPE(*) but you have to know, how to decode the structure in C. You can do it by reverse engineering.

There is actually a macro for the size of the structure in the referenced TS. See CFI_attribute_allocatable . But the problem with dereferencing remains. There is a standard array descriptor proposed, but this TS is not yet supported by some compilers (notably by gfortran).

What is possible, is to just pass the pointer to the structure to a generic function such as qsort . Then you use just a piece of memory, that happens to contain also some pointer or allocatable descriptor, but C does not have to know about them.

Another possibility is to construct a derived type which holds only type(c_ptr) , instead of Fortran pointers and use c_loc() to fill them. Be sure to have only contiguous arrays then.