指针到成员作为模板参数推导 [英] Pointer-to-member as template parameter deduction

问题描述

我想将指向成员的指针作为foo1的模板参数。这里是代码：

I want to get pointer-to-member as template parameter to the foo1. Here is code:

struct baz{
    int qux;
};

template<typename C, typename T, T C::*m>
struct foo1{};

template<typename C, typename T>
void barr2(T C::*m){
}

template<typename C, typename T>
void barr1(T C::*m){
    barr2(m); // ok
    foo1<C, T, &baz::qux> _; // ok
    foo1<C, T, m> f; // g++4.6.1 error here; how to pass 'm' correctly ?
}

int main(){
    barr1(&baz::qux);
}

那么它应该是什么样子？

So how it should look like?

推荐答案

它不适用于你，因为你试图在编译时表达式中使用运行时信息。它与使用从控制台读取的专门化模板的整数相同。

It doesn't work for you because you are trying to use run-time information in a compile-time expression. It is the same as using integer that you read from console to specialize a template. It is not meant to work.

这不一定能解决你的问题，但如果 barr1 函数是为了减轻打字负担，这样的东西可能为你工作：

It doesn't necessarily solve your problem, but if the intent of barr1 function was to ease typing burden, something like this may work for you:

struct baz{
    int qux;
};

template<typename C, typename T, T C::*m>
struct foo1 {};

#define FOO(Class, Member)                                  \
    foo1<Class, decltype(Class::Member), &Class::Member>

int main(){
    FOO(baz, qux) f;
}

这篇关于指针到成员作为模板参数推导的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！