在http://blogs.msdn.com/b/vcblog/archive/2011/09/12/10209291.aspx上,VC++团队正式声明他们尚未实现C++ 11核心功能"Expression SFINAE".但是,从http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2008/n2634.html复制的以下代码示例将被VC++编译器接受.
例1:
template <int I> struct A {};
char xxx(int);
char xxx(float);
template <class T> A<sizeof(xxx((T)0))> f(T){}
int main()
{
f(1);
}
例2:
struct X {};
struct Y
{
Y(X){}
};
template <class T> auto f(T t1, T t2) -> decltype(t1 + t2); // #1
X f(Y, Y); // #2
X x1, x2;
X x3 = f(x1, x2); // deduction fails on #1 (cannot add X+X), calls #2
我的问题是:什么是"表达SFINAE"？
如果基类没有提供方法,你将如何填写方法？如果提供的话,我想重用基类方法.
例如:
#include <iostream>
struct Base0 { };
struct Base1 { void m() { std::cout<<"Base1\n"; } };
template<typename T>
struct Derived : public T {
//if T doesn't provide m, define it here, otherwise reuse the base class method
void m(){ /*? std::cout<<"Derived\n"; ?*/ }
};
int main(){
Derived<Base0> d0;
d0.m(); //should print "Derived"
Derived<Base1> d1;
d1.m(); //should print "Base1"
}
如何在对数（至少以二为底）编译时间（严格来说，以对数实例化数量）中定义聚合的数量？
我目前能做的就是在线性时间内实现期望的目标：
#include <type_traits>
#include <utility>
struct filler { template< typename type > operator type (); };
template< typename A, typename index_sequence = std::index_sequence<>, typename = void >
struct aggregate_arity
: index_sequence
{
};
template< typename A, std::size_t ...indices >
struct aggregate_arity< A, std::index_sequence< indices... >, std::__void_t< decltype(A{(indices, std::declval< filler >())..., std::declval< filler >()}) > >
: aggregate_arity< A, std::index_sequence< indices..., sizeof...(indices) > >
{
};
struct A0 {};
struct A1 { double x; };
struct A2 { int i; …