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我正在尝试使用 SWIG 将一些使用 std::variant 的 C++17 代码包装在 Python 中。

我找到了这个关于包装 boost::variant 的答案（https://stackoverflow.com/a/58513139），并且我设法调整了代码，以便它可以与 std::variant 一起使用。但是，根据答案的规范，代码应该可以工作，以便“在 C++ 函数采用 boost::variant 的任何地方，我们都应该透明地接受该变体可以为该函数参数保存的任何类型”。似乎只有在使用对 std::variant 的 const 引用时才能满足此要求。例如，如果在我的 c++ 文件 dummy.cpp 我有

void foo(std::variant<int, double> value)
{
  std::cout << "foo" << std::endl;
}

void bar(const std::variant<int, double>& value)
{
  std::cout << "bar" << std::endl;
}

然后在我的 dummy.i

%include "variant.i" # My customization of boost_variant.i
%std_variant(DummyVariant, int, double); # Creating the variant type for SWIG/python

当我尝试从 Python 脚本中使用它时，以下工作

from dummy import bar
bar(5)

但是，这不会：

from dummy import foo
foo(5)

我得到：TypeError：在方法“foo”中，类型为“std::variant<int,double>”的参数 1。

有人知道我的 variant.i 文件中缺少什么以使其按预期工作吗？这是我拥有的文件：

%{
#include <variant>

static PyObject *this_module = NULL;
%}

%init %{
  // We need to "borrow" a reference to this for our typemaps to be able to look up the right functions
  this_module = m; // borrow should be fine since we can only get called when our module is loaded right?
  // Wouldn't it be nice if $module worked *anywhere*
%}

#define FE_0(...)
#define FE_1(action,a1) action(0,a1)
#define FE_2(action,a1,a2) action(0,a1); action(1,a2)
#define FE_3(action,a1,a2,a3) action(0,a1); action(1,a2); action(2,a3)
#define FE_4(action,a1,a2,a3,a4) action(0,a1); action(1,a2); action(2,a3); action(3,a4)
#define FE_5(action,a1,a2,a3,a4,a5) action(0,a1); action(1,a2); action(2,a3); action(3,a4); action(4,a5)

#define GET_MACRO(_1,_2,_3,_4,_5,NAME,...) NAME
%define FOR_EACH(action,...)
  GET_MACRO(__VA_ARGS__, FE_5, FE_4, FE_3, FE_2, FE_1, FE_0)(action,__VA_ARGS__)
%enddef

#define in_helper(num,type) const type & convert_type ## num () { return std::get<type>(*$self); }
#define constructor_helper(num,type) variant(const type&)

%define %std_variant(Name, ...)

%rename(Name) std::variant<__VA_ARGS__>;

namespace std {
  struct variant<__VA_ARGS__> {
    variant();
    variant(const std::variant<__VA_ARGS__>&);
    FOR_EACH(constructor_helper, __VA_ARGS__);
    int index();

    %extend {
      FOR_EACH(in_helper, __VA_ARGS__);
    }
  };
}

%typemap(out) std::variant<__VA_ARGS__> {
  // Make our function output into a PyObject
  PyObject *tmp = SWIG_NewPointerObj(&$1, $&1_descriptor, 0); // Python does not own this object...

  // Pass that temporary PyObject into the helper function and get another PyObject back in exchange
  const std::string func_name = "convert_type" + std::to_string($1.index());
  $result = PyObject_CallMethod(tmp, func_name.c_str(),  "");
  Py_DECREF(tmp);
}

%typemap(in) const std::variant<__VA_ARGS__>& (PyObject *tmp=NULL) {
  // I don't much like having to "guess" the name of the make_variant we want to use here like this...
  // But it's hard to support both -builtin and regular modes and generically find the right code.
  PyObject *helper_func = PyObject_GetAttrString(this_module, "new_" #Name );
  assert(helper_func);
  // TODO: is O right, or should it be N?
  tmp = PyObject_CallFunction(helper_func, "O", $input);
  Py_DECREF(helper_func);
  if (!tmp) SWIG_fail; // An exception is already pending

  // TODO: if we cared, we chould short-circuit things a lot for the case where our input really was a variant object
  const int res = SWIG_ConvertPtr(tmp, (void**)&$1, $1_descriptor, 0);
  if (!SWIG_IsOK(res)) {
    SWIG_exception_fail(SWIG_ArgError(res), "Variant typemap failed, not sure if this can actually happen");
  }
}

%typemap(freearg) const std::variant<__VA_ARGS__>& %{
  Py_DECREF(tmp$argnum);
%}

%enddef

这几乎是从 Flexo 的回答中逐字提取的，保存一些更改以使其与 std::variant 而不是 boost::variant 一起使用。

4

1 回答 1

0

避免在界面上使用模板并创建应用程序二进制接口 (ABI) 的最简单方法。在您的情况下，这对应于创建一个接口dummy.i

%module dummy

%{
#include "dummy.h"

void mywrap_foo_int(int i) {
  std::variant<int, double> tmp;
  tmp = i;
  foo(tmp);
}
void mywrap_foo_double(double d) {
  std::variant<int, double> tmp;
  tmp = d;
  foo(tmp);
}
%}

%include "variant.i"
%std_variant(DummyVariant, int, double);
%include "dummy.h"

很快，您将遇到模板和嵌套模板。这可以像您已经完成的那样以通用方式完成，它处理bar. 包装的任务基本对应于此。

于 2021-01-29T20:28:39.003 回答