与LuaBridge出口集装箱类

Question

我想将SRect和SRectVector从C++导出到Lua，但编译失败。做这件事的正确方法是什么？编译器：vs2019，vc++11操作系统：Win10 64

Push((遇到编译错误，

我认为这个论点只是SRectVector*，为什么编译器认为它是"std:：vector<SRect，std:：分配器&lt_Ty&gt>'？

class SRect{
public:
int left;
int top;
int right;
int     bottom;
SRect(int l, int t, int r, int b)
: left(l)
, top(t)
, right(r)
, bottom(b){}
//...
};

typedef std::vector<SRect>    SRectVector;

luabridge::getGlobalNamespace(L)
.beginClass <SRect>("SRect")
.addConstructor <void(*) (int, int, int, int)>()
.addProperty("left", &SRect::left)
//...
.endClass()
.beginClass <SRectVector>("SRectVector")
.addFunction("Push",
std::function <void(SRectVector*, const SRect&)>(
[](SRectVector* vec, const SRect& rc) { (*vec).push_back(rc); }))
//...
.endClass()
.endNamespace();

```

1>E:CodeincludeLuaBridge/detail/TypeList.h(177): error C2664: 'luabridge::detail::TypeListValues<luabridge::detail::TypeList<Param,luabridge::detail::TypeList<const SRect&,luabridge::detail::MakeTypeList<>::Result>>>::TypeListValues(luabridge::detail::TypeListValues<luabridge::detail::TypeList<Param,luabridge::detail::TypeList<const SRect&,luabridge::detail::MakeTypeList<>::Result>>> &&)': cannot convert argument 1 from 'std::vector<SRect,std::allocator<_Ty>>' to 'Head'
1>        with
1>        [
1>            Param=SRectVector *
1>        ]
1>        and
1>        [
1>            _Ty=SRect
1>        ]
1>        and
1>        [
1>            Head=SRectVector *
1>        ]
1>E:CodeincludeLuaBridge/detail/TypeList.h(179): note: No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
1>E:CodeincludeLuaBridge/detail/TypeList.h(176): note: while compiling class template member function 'luabridge::detail::ArgList<Params,1>::ArgList(lua_State *)'

Answer 1

经过更多的挖掘，我找到了原因。一个完整的用户定义类可以很容易地导出。但容器指针不是。在我添加这样的代码后编译还可以，

namespace LuaBridge
{
template <>
struct Stack <SRectVector*>
{
static void push(lua_State* L, SRectVector* ptr)
{
SRectVector** pp = (SRectVector**)lua_newuserdata(L, sizeof(SRectVector*));
*pp = ptr;
}
static SRectVector* get(lua_State* L, int index)
{
return (SRectVector*)lua_touserdata(L, index);
}
};
}

或者添加一个更通用的

template <class T>
struct Stack <std::vector<T>*>
{
typedef typename std::vector<T>* ContainerPointerType;
static void push(lua_State* L, ContainerPointerType ptr)
{
ContainerPointerType* pp = (ContainerPointerType*)lua_newuserdata(L, sizeof(ContainerPointerType));
*pp = ptr;
}
static ContainerPointerType get(lua_State* L, int index)
{
return (ContainerPointerType)lua_touserdata(L, index);
}
};

Answer 2

不包括<LuaBridge/Vvector.h>，然后解决这个问题。