在Boost.MPL中已经实现了类似的东西，它被称为BOOST_MPL_HAS_XXX_TRAIT_DEF。请参阅：

There is something similar already implemented in Boost.MPL, it is called "BOOST_MPL_HAS_XXX_TRAIT_DEF". See:

http://www.boost.org/doc/libs/1_41_0/libs/mpl/doc/refmanual/has-xxx-trait-def.html

它可以检测类是否具有给定的类型。

It can detect if the class have a given named type.

对于你的具体情况，而不是传递函数指针作为参数（void（_t :: *）（）），尝试使用它在你的方法的主体，即像：

Also, for your specific case, instead of passing the function pointer as a parameter (void (_t::*)()), try to use it in the body of your method, i.e., something like:

template < typename _t >
static big test( sfinae<_t> )
{
  &_t::foo<_n>;
}