Question

如何使用SWIG包装此内容？

usecase5.h

#ifndef __USECASE5_H__
#define __USECASE5_H__

#include <math.h>

namespace foo_namespace {
    struct vec2
    {
        int x, y;

        vec2() {}

        explicit vec2( int a, int b )
        {
            x = a;
            y = b;
        }

        vec2 & operator =( vec2  const & v ) { x = v.x;  y = v.y;  return *this; }
        vec2 & operator/=( vec2  const & v ) { x /= v.x; y /= v.y; return *this; }

        vec2 xx() const { return vec2(x,x); }
    }
}

#endif

usecase6.h

#ifndef __USECASE6_H__
#define __USECASE6_H__

#include "usecase5.h"

namespace foo_namespace {
    vec2  usecase6_f1(const vec2 & x);
}

#endif

usecase6.cpp

namespace foo_namespace {
    vec2  usecase6_f1(const vec2 & x)
    {
        x = vec2();
    }
}

example.i

// GENERATED BY gen_swig.py at 2016-10-04 21:46

%module example

%{
//#include "usecase1.h"
//#include "usecase2.h"
//#include "usecase3.h"
//#include "usecase4.h"
#include "usecase5.h"
#include "usecase6.h"

using namespace foo_namespace;
%}

//%include "usecase1.h"
//%include "usecase2.h"
//%include "usecase3.h"
//%include "usecase4.h"
%include "usecase5.h"
%include "usecase6.h"

用swig -python -c++ example.i尝试了那个天真的example.i之后我会得到：

usecase5.h(19) : Warning 362: operator= ignored
usecase5.h(24) : Error: Syntax error in input(3).

那么，我怎么能包装这个小的虚拟c ++示例？

Answer 1

似乎这不是与swig相关的任何问题，但更多关于c ++方面，修复它我只需要修复c ++问题：

usecase5.h

#ifndef __USECASE5_H__
#define __USECASE5_H__

#include <math.h>

namespace foo_namespace {
    struct vec2
    {
        int x, y;

        vec2() {}

        explicit vec2( int a, int b )
        {
            x = a;
            y = b;
        }

        vec2 & operator =( vec2  const & v ) { x = v.x;  y = v.y;  return *this; }
        vec2 & operator/=( vec2  const & v ) { x /= v.x; y /= v.y; return *this; }

        vec2 xx() const { return vec2(x,x); }
    };
}

#endif

usecase6.cpp

#include "usecase5.h"

namespace foo_namespace {
    vec2  usecase6_f1(const vec2 & x)
    {
        return x;
    }
}

之后，包装器将成功构建并在python端正常工作。

包装仅在带有SWIG的标头中定义的命名空间自定义类型

1 个答案: