C++ 泛型成员函数指针作为模板参数
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原文地址: http://stackoverflow.com/questions/9779105/
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generic member function pointer as a template parameter
提问by Lorenzo Pistone
Consider this code:
考虑这个代码：
#include <iostream>
using namespace std;
class hello{
public:
void f(){
cout<<"f"<<endl;
}
virtual void ff(){
cout<<"ff"<<endl;
}
};
#define call_mem_fn(object, ptr) ((object).*(ptr))
template<R (C::*ptr_to_mem)(Args...)> void proxycall(C& obj){
cout<<"hello"<<endl;
call_mem_fn(obj, ptr_to_mem)();
}
int main(){
hello obj;
proxycall<&hello::f>(obj);
}
Of course this won't compile at line 16, because the compiler doesn't know what R, Cand Args, are. But there's another problem: if one tries to define those template parameters right before ptr_to_mem, he runs into this bad situation:
当然，这不会在第 16 行编译，因为编译器不知道R、C和Args是什么。但是还有另一个问题：如果试图在之前定义这些模板参数ptr_to_mem，他会遇到这种糟糕的情况：
template<typename R, typename C, typename... Args, R (C::*ptr_to_mem)(Args...)>
// ^variadic template, but not as last parameter!
void proxycall(C& obj){
cout<<"hello"<<endl;
call_mem_fn(obj, ptr_to_mem)();
}
int main(){
hello obj;
proxycall<void, hello, &hello::f>(obj);
}
Surprisingly, g++ does not complain about Argsnot being the last parameter in the template list, but anyway it cannot bind proxycallto the right template function, and just notes that it's a possible candidate.
令人惊讶的是，g++ 并没有抱怨它不是Args模板列表中的最后一个参数，但无论如何它无法绑定proxycall到正确的模板函数，并且只是指出它是一个可能的候选者。
Any solution? My last resort is to pass the member function pointer as an argument, but if I could pass it as a template parameter it would fit better with the rest of my code.
有什么解决办法吗？我最后的手段是将成员函数指针作为参数传递，但如果我可以将它作为模板参数传递，它会更适合我的其余代码。
EDIT: as some have pointed out, the example seems pointless because proxycall isn't going to pass any argument. This is not true in the actual code I'm working on: the arguments are fetched with some template tricks from a Lua stack. But that part of the code is irrelevant to the question, and rather lengthy, so I won't paste it here.
编辑：正如一些人指出的那样，这个例子似乎毫无意义，因为 proxycall 不会传递任何参数。这在我正在处理的实际代码中并非如此：参数是通过 Lua 堆栈中的一些模板技巧获取的。但是那部分代码与问题无关，而且相当冗长，所以我不会在这里粘贴。
回答by Kerrek SB
You could try something like this:
你可以尝试这样的事情：
template <typename T, typename R, typename ...Args>
R proxycall(T & obj, R (T::*mf)(Args...), Args &&... args)
{
return (obj.*mf)(std::forward<Args>(args)...);
}
Usage: proxycall(obj, &hello::f);
用法： proxycall(obj, &hello::f);
Alternatively, to make the PTMF into a template argument, try specialization:
或者，要使 PTMF 成为模板参数，请尝试特化：
template <typename T, T> struct proxy;
template <typename T, typename R, typename ...Args, R (T::*mf)(Args...)>
struct proxy<R (T::*)(Args...), mf>
{
static R call(T & obj, Args &&... args)
{
return (obj.*mf)(std::forward<Args>(args)...);
}
};
Usage:
用法：
hello obj;
proxy<void(hello::*)(), &hello::f>::call(obj);
// or
typedef proxy<void(hello::*)(), &hello::f> hello_proxy;
hello_proxy::call(obj);
