我想区分C ++模板结构中的重载成员函数。如果此方法存在,则来自专用struct get_pointer
的静态方法distinguish_foo
应返回指向derived::foo
的指针。否则我想返回指向pseudo::foo
的指针
我怎样才能做到这一点?
struct sample
{
void foo() { std::cout << "void sample::foo()" << std::endl; }
void foo(int) { std::cout << "void sample::foo(int)" << std::endl; }
void foo(int, int) { std::cout << "void sample::foo(int, int)" << std::endl; }
void foo(int, int, int) { std::cout << "void sample::foo(int, int, int)" << std::endl; }
};
struct other
{
void foo() { std::cout << "void other::foo()" << std::endl; }
void foo(int) { std::cout << "void other::foo(int)" << std::endl; }
void foo(int, int) { std::cout << "void other::foo(int, int)" << std::endl; }
void foo(int, int, int) { std::cout << "void other::foo(int, int, int)" << std::endl; }
};
struct derived : sample, other
{
using sample::foo;
using other::foo;
};
template<typename signature>
struct distinguish_foo;
template<class C, typename R>
struct distinguish_foo<R (C::*)()>
{
typedef char(&unique)[0xffff];
struct pseudo { unique foo(); };
struct host : C, pseudo {};
template<typename R>
static R (host::*get_pointer())() { return &host::foo; }
};
template<class C, typename R, typename P0>
struct distinguish_foo<R (C::*)(P0)>
{
typedef char(&unique)[0xffff];
struct pseudo { unique foo(P0); };
struct host : C, pseudo {};
template<typename R>
static R (host::*get_pointer())(P0) { return &host::foo; }
};
template<class C, typename R, typename P0, typename P1>
struct distinguish_foo<R (C::*)(P0, P1)>
{
typedef char(&unique)[0xffff];
struct pseudo { unique foo(P0, P1); };
struct host : C, pseudo {};
template<typename R>
static R (host::*get_pointer())(P0, P1) { return &host::foo; }
};
template<class C, typename R, typename P0, typename P1, typename P2>
struct distinguish_foo<R (C::*)(P0, P1, P2)>
{
typedef char(&unique)[0xffff];
struct pseudo { unique foo(P0, P1, P2); };
struct host : C, pseudo {};
template<typename R>
static R (host::*get_pointer())(P0, P1, P2) { return &host::foo; }
};