我正在尝试，但我不知道当我将派生类的函数成员指针作为模板参数传递时如何(但可能不可能)捕获相同类型的类。
例如，有这段代码，
class CObject{
public:
string m_id;
void setID(const char * id){
printf("assign id %s\n",id);
m_id=string(id);
}
~CObject(){}
};
class CDerivedObject:public CObject{
public:
~CDerivedObject(){};
};
有了这个函数模板，
template < typename _R, class _T, typename..._A>
void function_template_test(_R (_T:: *function_type)(_A...) )
{
printf("ObjectTypeIs: %s\n",typeid( _T).name());
}
在主函数中，我调用 function_template_test 传递函数指针成员 CDerivedObject::setID，
function_template_test(&CDerivedObject::setID);
如果我用 g++ 编译此代码(我的最新版本是 5.4.0)，它会打印 7CObject ，这意味着 _T 是 CObject 类型，但在参数中我传递了 CDerivedObject::setID。我知道函数setID函数属于CObject，所以编译器以这种方式获取类型，但是...
有没有可能的方法来捕获我传递给模板参数的类类型(即 _T 是 CDerivedObject)？
编辑 1
我的问题的解决方案可能是这样的，
template < class _C, class _R, typename _T, typename..._A>
void _function_template_test(_R (_T:: *function_type)(_A...) )
{
printf("ObjectTypeIs: %s\n",typeid( _C).name());
}
#define function_template_test(cl, fun)\
_function_template_test<cl>(&cl::fun)
然后我将该函数称为，
function_template_test(CDerivedObject,setID);
是一个解决方案，但我更喜欢模板而不是宏。
最佳答案
考虑这个程序:
struct Base { void foo() {} };
struct Derived: Base {};
auto main() -> int
{
int x = &Derived::foo;
}
这是使用 MinGW g++ 编译的结果:
[H:\forums\so\056]
> g++ main.cpp
main.cpp: In function 'int main()':
main.cpp:6:23: error: cannot convert 'void (Base::*)()' to 'int' in initialization
int x = &Derived::foo;
^~~
[H:\forums\so\056]
> _
As you can see the type of &Derived::foo is void (Base::*)() (and yes, that's according to the standard).
I.e., when you ask
” Is there a possible way to capture the class type I'm passing on template parameter
... that's what you're doing.
To make that type void (Derived::*)() you need a function declared in Derived.
For your case of calling a function that takes a member function pointer, you can alternatively specify all the template arguments explicitly. That works because a void (Base::*)() converts implicitly to void (Derived::*)(). And yes that's counter-intuitive, and constitutes a little loop-hole in the C++ type system, but it's the way the logic works out when you consider what kinds of objects a member function can be called on.
Re the loop-hole, here's an example (see ma, no casts!):
#include <iostream>
#include <stack>
using namespace std;
auto main()
-> int
{
stack<int> st;
for( int const i : {1, 2, 3} ) { st.push( i ); }
struct Hack: stack<int> { using stack<int>::c; };
int const n = (st.*&Hack::c).size();
cout << "That stack contains " << n << " items." << endl;
}
关于c++ - 获取与模板 arg 中传递的函数成员类型相同的类，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/48601298/