[英]Export container class with LuaBridge
我想将 SRect 和 SRectVector 从 C++ 导出到 Lua,但编译失败。 什么是正确的方法呢? 编译器:vs2019,vc++11 操作系统:Win10 64
push() 遇到编译错误,
我认为参数只是 SRectVector *,为什么编译器认为它是 'std::vector<SRect,std::allocator<_Ty>>'?
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:\Code\include\LuaBridge/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:\Code\include\LuaBridge/detail/TypeList.h(179): note: No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
1>E:\Code\include\LuaBridge/detail/TypeList.h(176): note: while compiling class template member function 'luabridge::detail::ArgList<Params,1>::ArgList(lua_State *)'
经过更多的挖掘,我找到了原因。 可以轻松导出完整的用户定义类。 但容器指针不是。 添加这样的代码后编译正常,
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);
}
};
不要包含<LuaBridge/Vector.h>,然后解决这个问题。
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