C ++为什么SFINAE仅使用类模板参数会失败？ [英] C++ why does SFINAE fail with only a class template parameter?

问题描述

我正在使用此答案样式的SFINAE，以便通过使用适当的方法来调用通用矢量对象成员函数。例如，以下代码首先调用 operator []（int）const ，如果不存在，则调用 operator（）（int）const ：

I'm using SFINAE in the style of this answer in order to call a generic vector object by using an appropriate member function. For example, the following code calls operator[](int) const first, and if that doesn't exist then operator()(int) const:

template<int I> struct rank : rank<I-1> { static_assert(I > 0, ""); };
template<> struct rank<0> {};

template<typename VectorType>
struct VectorWrapper
{
    auto get(int i) const
    {
        return get(v, i, rank<5>());
    }

    template<typename V, typename = std::enable_if_t<has_bracket_operator<const V>::value> >
    auto get(V const& v, int i, rank<2>) const
    {
        return v[i];
    }

    template<typename V, typename = std::enable_if_t<has_parenthesis_operator<const V>::value> >
    auto get(V const& v, int i, rank<1>) const
    {
        return v(i);
    }

    VectorType v;
};

使用 has_bracket_operator 和 has_parenthesis_operator 特征已按照此线程，整个编译并似乎可以正常工作。

With the has_bracket_operator and has_parenthesis_operator traits set up as suggested in this thread, the whole compiles and seems to work.

但是，从一开始就没有必要将成员向量传递给重载的类模板，因此我尝试在不传递的情况下进行设置。为此，我将模板参数 V 替换为用于设置类模板的 VectorType 参数：

However, passing the member vector to the overloaded class templates seems unnecessary from the first, so I tried to set up the same without passing it. For this, I replaced the template parameter V with the VectorType parameter used to set up the class template:

    template<typename = std::enable_if_t<has_bracket_operator<VectorType>::value> >
    auto get(int i, rank<2>) const
    {
        return v[i];
    }

    template<typename = std::enable_if_t<has_parenthesis_operator<VectorType>::value> >
    auto get(int i, rank<1>) const
    {
        return v(i);
    }

现在，但是编译失败（在gcc 5.1.0中）出现以下错误消息：

Now, however, the compilation fails (in gcc 5.1.0) with the following error message:

/usr/local/include/c++/5.1.0/type_traits: In substitution of 'template<bool _Cond, class _Tp> using enable_if_t = typename std::enable_if::type [with bool _Cond = has_parenthesis_operator<std::vector<int> >::value; _Tp = void]':
main.cpp:46:10:   required from 'struct VectorWrapper<std::vector<int> >'
main.cpp:59:38:   required from here
/usr/local/include/c++/5.1.0/type_traits:2388:61: error: no type named 'type' in 'struct std::enable_if<false, void>'
     using enable_if_t = typename enable_if<_Cond, _Tp>::type;

演示

问题：

• 此编译错误的原因是什么？


• 除了第一个代码块以外，是否还有其他适当的解决方法？ （也就是说，它保留了通常的编码样式-无需传递成员。）

推荐答案

SFINAE来自[temp.deduct] / 8，重点是我：

SFINAE comes to us from [temp.deduct]/8, emphasis mine:

如果替换导致类型或表达式无效，类型推导失败。无效的类型或表达式是
，如果使用替换的参数编写的话，它将格式错误，并且需要诊断。 [注意：
如果不需要诊断，则程序仍然格式错误。访问检查是替换
过程的一部分。 —end note] 仅在函数类型的直接上下文中无效的类型和表达式，以及其模板参数类型的
会导致推论失败。

If a substitution results in an invalid type or expression, type deduction fails. An invalid type or expression is one that would be ill-formed, with a diagnostic required, if written using the substituted arguments. [ Note: If no diagnostic is required, the program is still ill-formed. Access checking is done as part of the substitution process. —end note ] Only invalid types and expressions in the immediate context of the function type and its template parameter types can result in a deduction failure.

直接上下文是模板声明中的内容。在您的第一个示例中：

The immediate context is what's in the template declaration. In your initial example:

template<typename V, typename = std::enable_if_t<has_bracket_operator<const V>::value> >
auto get(V const& v, int i, rank<2>) const

V 是即时上下文，因此 enable_if 的替换失败只是推论失败。

V is in the immediate context, so a substitution failure on the enable_if is just a deduction failure.

但是，在第二个示例中：

However, in your second example:

template<typename = std::enable_if_t<has_bracket_operator<VectorType>::value> >
auto get(int i, rank<2>) const

VectorType 不在 get 的直接上下文中，因此此处的失败不是推论失败，而是硬错误。

VectorType is not in the immediate context of get, so a failure here would not be a deduction failure, it would be a hard error.

除非 VectorType 恰好具有所有这些运算符。

Unless VectorType happens to have all of these operators.

解决任何模板问题的方法是仅添加更多模板。在这种情况下，通过引入另一种类型来强制 VectorType 处于直接上下文中：

The solution to any template problem is to just add more template. In this case, force VectorType to be in the immediate context by introducing another type:

template<typename T=VectorType, typename = std::enable_if_t<has_bracket_operator<T>::value> >
auto get(int i, rank<2>) const

并调用 get<>（）。

这篇关于C ++为什么SFINAE仅使用类模板参数会失败？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！