在void *和指向成员函数的指针之间进行转换 [英] Casting between void * and a pointer to member function

问题描述

我目前使用GCC 4.4，我有相当的头痛转换void *和指向成员函数的指针。我试图写一个易于使用的库来绑定C ++对象到Lua解释器，如下：

I'm currently using GCC 4.4, and I'm having quite the headache casting between void * and a pointer to member function. I'm trying to write an easy-to-use library for binding C++ objects to a Lua interpreter, like so:

LuaObject<Foo> lobj = registerObject(L, "foo", fooObject);
lobj.addField(L, "bar", &Foo::bar);

我已经做了大部分工作，除了下面的功能函数签名，直到我有机会概括它）：

I've got most of it done, except for the following function (which is specific to a certain function signature until I have a chance to generalize it):

template <class T>
int call_int_function(lua_State *L) 
{
    // this next line is problematic
    void (T::*method)(int, int) = reinterpret_cast<void (T::*)(int, int)>(lua_touserdata(L, lua_upvalueindex(1)));
    T *obj = reinterpret_cast<T *>(lua_touserdata(L, 1));

    (obj->*method)(lua_tointeger(L, 2), lua_tointeger(L, 3));
    return 0;
}

对于不熟悉Lua的人， lua_touserdata L，lua_upvalueindex（1））获取与闭包相关联的第一个值（在这种情况下，它是成员函数的指针），并将其作为void * 。 GCC抱怨void * - > void（T :: *）（int，int）是无效的转换。关于如何解决这个问题的任何想法？

For those of you unfamiliar with Lua, lua_touserdata(L, lua_upvalueindex(1)) gets the first value associated with a closure (in this case, it's the pointer to member function) and returns it as a void *. GCC complains that void * -> void (T::*)(int, int) is an invalid cast. Any ideas on how to get around this?

推荐答案

您无法将指向成员的指针转换为 void * 或任何其他常规指针类型。指针到成员不是常规指针的地址。你最有可能需要做的是将成员函数包装在一个常规函数中。 C ++常见问题解答版本对此进行了详细说明。主要问题是实现指针到成员所需的数据不只是一个地址，实际上不同

You cannot cast a pointer-to-member to void * or to any other "regular" pointer type. Pointers-to-members are not addresses the way regular pointers are. What you most likely will need to do is wrap your member function in a regular function. The C++ FAQ Lite explains this in some detail. The main issue is that the data needed to implement a pointer-to-member is not just an address, and in fact varies tremendously based on the compiler implementation.

我假设你可以控制用户数据 lua_touserdata 正在返回。它不能是指向成员的指针，因为没有合法的方式来获取此信息。但你还有其他选择：

I presume you have control over what the user data lua_touserdata is returning. It can't be a pointer-to-member since there isn't a legal way to get this information back out. But you do have some other choices:

• 最简单的选择是将你的成员函数包装在一个自由函数中， 。该自由函数应该将对象作为其第一个参数。请参阅下面的代码示例。

• The simplest choice is probably to wrap your member function in a free function and return that. That free function should take the object as its first argument. See the code sample below.

使用类似于 Boost.Bind的mem_fun 来返回一个函数对象，你可以正确的模板。我看不到这更容易，但它会让你关联更多的状态与函数return如果你需要。

Use a technique similar to that of Boost.Bind's mem_fun to return a function object, which you can template on appropriately. I don't see that this is easier, but it would let you associate the more state with the function return if you needed to.

这里是使用第一种方式重写你的函数：

Here's a rewrite of your function using the first way:

template <class T>
int call_int_function(lua_State *L) 
{
    void (*method)(T*, int, int) = reinterpret_cast<void (*)(T*, int, int)>(lua_touserdata(L, lua_upvalueindex(1)));
    T *obj = reinterpret_cast<T *>(lua_touserdata(L, 1));

   method(obj, lua_tointeger(L, 2), lua_tointeger(L, 3));
   return 0;
}

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