Question

我有一个结构：

struct some_struct_s {
   int arg1;
   int arg2;
};

我有一个C函数：

int func(some_struct_s *output);

两个人都%included进入了我的SWIG文件。

我希望将some_struct_s *output视为输出参数。 Python示例：

int_val, some_struct_output = func()

POD类型（第10.1.3节）中介绍了“输出参数”，但非POD类型则没有。

如何告诉SWIG我希望some_struct_s *output作为输出参数？

Answer 1

来自documentation：

11.5.7“ argout”类型映射

“ argout”类型映射用于从参数返回值。这最常用于为需要返回多个值的C / C ++函数编写包装器。几乎总是将“ argout”类型映射与“ in”类型映射组合---可能会忽略输入值。...

这是您代码的完整示例（为简洁起见，没有错误检查）：

%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;
}

%}

下面的演示。请注意，int返回值为零以及输出参数为元组作为列表返回。

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

如果您不想使用类型映射，还可以注入代码来创建对象，然后将其返回以对用户隐藏：

%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
%}

演示：

>>> 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

如果仔细选择SWIG宏，则可以使类型映射语言不可知：

%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 {
    %append_output(SWIG_NewPointerObj($1,$1_descriptor,1));
}

%inline %{

struct some_struct_s {
   int arg1;
   int arg2;
};

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

%}

演示：

>>> 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

SWIG结构指针作为输出参数

1 个答案: