如何使用 sfinae 选择构造函数?
在模板元编程中,可以在返回类型上使用SFINAE来选择某个模板成员函数,即
In template meta programming, one can use SFINAE on the return type to choose a certain template member function, i.e.
template<int N> struct A {
int sum() const noexcept
{ return _sum<N-1>(); }
private:
int _data[N];
template<int I> typename std::enable_if< I,int>::type _sum() const noexcept
{ return _sum<I-1>() + _data[I]; }
template<int I> typename std::enable_if<!I,int>::type _sum() const noexcept
{ return _data[I]; }
};
但是,这不适用于构造函数.假设,我想声明构造函数
However, this doesn't work on constructors. Suppose, I want to declare the constructor
template<int N> struct A {
/* ... */
template<int otherN>
explicit(A<otherN> const&); // only sensible if otherN >= N
};
但不允许 otherN <N
.
那么,可以在这里使用 SFINAE 吗?我只对允许自动模板参数推导的解决方案感兴趣,所以
So, can SFINAE be used here? I'm only interested in solutions which allow automatic template-parameter deduction, so that
A<4> a4{};
A<5> a5{};
A<6> a6{a4}; // doesn't compile
A<3> a3{a5}; // compiles and automatically finds the correct constructor
注意:这是一个非常简化的示例,其中 SFINAE 可能有点矫枉过正,而 static_assert
可能就足够了.但是,我想知道我是否可以改用 SFINAE.
Note: this is a very simplified example where SFINAE may be overkill and static_assert
may suffice. However, I want to know whether I can use SFINAE instead.
推荐答案
您可以向模板添加默认类型参数:
You can add a defaulted type argument to the template:
template <int otherN, typename = typename std::enable_if<otherN >= N>::type>
explicit A(A<otherN> const &);
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