SWIG struct pointer as output parameter

Question

I have a struct:

struct some_struct_s {
   int arg1;
   int arg2;
};

I have a C function:

int func(some_struct_s *output);

Both are %included into my SWIG file.

I want some_struct_s *output to be treated like an output parameter. Python example:

int_val, some_struct_output = func()

“Output parameters” is covered in the manual for POD-types (sec 10.1.3), but not for non-POD types.

How do I tell SWIG I want some_struct_s *output to be an output parameter?

Asked By: Michael Labbé

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Source

Answer 1

From the documentation:

11.5.7 "argout" typemap

The "argout" typemap is used to return values from arguments. This is most commonly used to write wrappers for C/C++ functions that need to return multiple values. The "argout" typemap is almost always combined with an "in" typemap—possibly to ignore the input value….

Here’s a complete example for your code (no error checking for brevity):

%module test

// Declare an input typemap that suppresses requiring any input and
// declare a temporary stack variable to hold the return data.
%typemap(in,numinputs=0) some_struct_s* (some_struct_s tmp) %{
    $1 = &tmp;
%}

// Declare an output argument typemap.  In this case, we'll use
// a tuple to hold the structure data (no error checking).
%typemap(argout) some_struct_s* (PyObject* o) %{
    o = PyTuple_New(2);
    PyTuple_SET_ITEM(o,0,PyLong_FromLong($1->arg1));
    PyTuple_SET_ITEM(o,1,PyLong_FromLong($1->arg2));
    $result = SWIG_Python_AppendOutput($result,o);
%}

// Instead of a header file, we'll just declare this code inline.
// This includes the code in the wrapper, as well as telling SWIG
// to create wrappers in the target language.
%inline %{

struct some_struct_s {
   int arg1;
   int arg2;
};

int func(some_struct_s *output) {
    output->arg1 = 1;
    output->arg2 = 2;
    return 0;
}

%}

Demo below. Note that the int return value of zero as well as the output parameter as a tuple are returned as a list.

>>> import test
>>> test.func()
[0, (1, 2)]

If you don’t want typemaps, you can also inject code to create the object and return it to hide it from the user:

%module test

%rename(_func) func; // Give the wrapper a different name

%inline %{

struct some_struct_s {
   int arg1;
   int arg2;
};

int func(struct some_struct_s *output)
{
    output->arg1 = 1;
    output->arg2 = 2;
    return 0;
}

%}

// Declare your interface
%pythoncode %{
def func():
    s = some_struct_s()
    r = _func(s)
    return r, s
%}

Demo:

>>> import test
>>> r, s = test.func()
>>> r
0
>>> s
<test.some_struct_s; proxy of <Swig Object of type 'some_struct_s *' at 0x000001511D70A880> >
>>> s.arg1
1
>>> s.arg2
2

You can make the typemap language agnostic if you carefully select SWIG macros:

%module test

%typemap(in,numinputs=0) struct some_struct_s *output %{
    $1 = malloc(sizeof(struct some_struct_s));
%}

%typemap(argout) struct some_struct_s* output {
    // The last parameter passes ownership of the pointer
    // to Python so it will be freed when the object's
    // reference count goes to zero.
    %append_output(SWIG_NewPointerObj($1, $1_descriptor, SWIG_POINTER_OWN));
}

%inline %{

struct some_struct_s {
   int arg1;
   int arg2;
};

int func(struct some_struct_s *output)
{
    output->arg1 = 1;
    output->arg2 = 2;
    return 0;
}

%}

Demo:

>>> import test
>>> r, s = test.func()
>>> r
0
>>> s
<test.some_struct_s; proxy of <Swig Object of type 'some_struct_s *' at 0x000001DD0425A700> >
>>> s.arg1
1
>>> s.arg2
2

Answered By: Mark Tolonen

SWIG struct pointer as output parameter

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