如何使用sfinae选择构造函数? [英] How to use sfinae for selecting constructors?
问题描述
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,例如
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,这对构造函数不起作用。假设,我想声明构造函数
template< int N& struct A {
/ * ... * /
template< int otherN>
explicit(A< otherN> const&); // only sensible if otherN> = N
};
但不允许 otherN<因此,可以在这里使用SFINAE吗?我只对允许自动模板的解决方案感兴趣。 - 参数扣除,使得
A 4 a4 {};
A< 5> a5 {};
A< 6> a6 {a4}; //不编译
A< 3> a3 {a5}; //编译并自动找到正确的构造函数
注意:这是一个非常简化的示例, SFINAE可能过分,并且 static_assert
可能就足够了。但是,我想知道我是否可以使用SFINAE。
You可以为模板添加默认类型参数:
template< int otherN,typename = typename std :: enable_if< otherN> ; = N> :: type>
explicit A(A< otherN> const&);
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]; }
};
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
};
but disallow it for otherN < N
.
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
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.
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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