将函数 ptr 获取到实例化类的成员函数？
class gfx {
void resize(int x, int y);
}
gfx g;
我可以以某种方式将 g.resize 转换为 'void (*)(int, int)' 吗？
class gfx {
void resize(int x, int y);
}
gfx g;
can i cast g.resize to a 'void (*)(int, int)' somehow?
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不。
gfx::resize的类型为void(gfx::*)(int, int)。您无法将其有意义地转换为void(*)(int, int)类型，因为您只能将其作为gfx类型的对象上的成员函数进行调用。成员函数只能在类的实例上调用，因此给定您的
gfx g;，您可以调用g.resize()，但您不能仅仅像调用普通函数一样调用resize()。普通函数指针不能指向成员函数，因为它没有任何方法将函数调用绑定到类的实例。从概念上讲，成员函数需要一个附加参数，即
this参数，该参数指向调用它的类的实例。如果您希望能够通过普通函数指针调用成员函数，则可以创建一个带有参数的非成员函数（或静态成员函数）包装器，您可以向该参数传递要调用该成员的对象的实例功能。例如，您可以：
此非成员
resize函数的类型为void(*)(gfx*, int, int)并且可以作为非成员函数进行调用。这种方法的通用形式是 C++0x 中的std::function和std::bind设施（您也可以在 Boost 和 C++ TR1）。No.
gfx::resizeis of typevoid(gfx::*)(int, int). You can't meaningfully convert it to typevoid(*)(int, int)because you can only call it as a member function on an object of typegfx.A member function can only be called on an instance of the class, so given your
gfx g;, you can callg.resize(), but you can't just callresize()like it were an ordinary function. An ordinary function pointer can't point to a member function because it doesn't have any way to bind the function call to an instance of the class.Conceptually, a member function takes an additional parameter, the
thisparameter, that points to the instance of the class on which it was called.If you want to be able to call a member function via an ordinary function pointer, you can create a nonmember function (or a static member function) wrapper with a parameter to which you can pass the instance of the object on which to call the member function. For example, you could have:
This nonmember
resizefunction has typevoid(*)(gfx*, int, int)and can be called as a nonmember function. The generalized form of this approach is thestd::functionandstd::bindfacilities found in C++0x (you can also find them in Boost and in C++ TR1).是的，你可以这样做：
Yes, you can do it: