模板推演失败

Question

我正在尝试从 boost 库中实现绑定功能。下面你可以看到定义了 operator().

的主结构 bind_t

我的问题如下：为什么我们要在返回类型operator()中的decltype中明确指定返回类型call()作为成员函数（如果我在[=之前删除this-> 16=], g++中模板实参推导失败。)

同样有趣的是，使用 clang++ 就没有这样的问题。

我不知道为什么会这样。

template <typename F, typename  ... P>
struct bind_t {
private:
    std::tuple<typename holder<P>::type...> p;
    F func;
    template <size_t ... N, typename ... Args>
    auto call(index_list<N ...>, Args const& ... args) const -> decltype(func(std::get<N>(p)(args...)...)) {
        return func(std::get<N>(p)(args...)...);
    }
public:
    bind_t(F f, P ... p):
        p(std::move(p)...),
        func(std::move(f))
    {}
    template <typename ... Args>
    auto operator()(Args const& ... args) const -> decltype(this->call(typename indices_by_num<sizeof...(P)>::type(), args...)) {
        typename indices_by_num<sizeof...(P)>::type indices;
        return call(indices, args...);
    }
};

full source 实施
simple usecase

Answer 1

这是一个 gcc 错误，记录在错误报告 decltype needs explicit 'this' pointer in member function declaration of template class with trailing return type 中说：

When using trailing return-type for member functions of a template class, the 'this' pointer must be explicitly mentioned. This should not be necessary (The implicit 'this' works with a non-template class).

Example:
template <typename T>
struct DecltypeConstThis {

    T f() const { return T{}; }

    auto g() -> decltype(this->f()) { return this->f(); }
    auto h() const  ->  decltype(f()) { return f(); } // this should work the same as g() above (with implicit 'this')

};

struct Working {
    int f() const { return 0; }
    auto h() const -> decltype(f()) { return 0; }
};


int main() {

    Working w;
    w.h();

    DecltypeConstThis<int> d;
    d.g();
    d.h();

    return 0;
}

该报告被标记为已修复，看起来这项工作开始在 gcc 5.1 中工作 (see it live)。

模板推演失败

Template deduction failure

c++

gcc

templates

sfinae

variadic-templates