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模板转换运算符的分辨率不明确

conversion-operator language-lawyer templates c++

6

Guillaume Racicot · 技术社区 · 7 年前

我不得不做一个类似的代码:

#include <type_traits>

template<typename S>
struct probe {
    template<typename T, typename U = S, std::enable_if_t<
        std::is_same<T&, U>::value &&
        !std::is_const<T>::value, int> = 0>
    operator T& () const;

    template<typename T, typename U = S&&, std::enable_if_t<
        std::is_same<T&&, U>::value &&
        !std::is_const<T>::value, int> = 0>
    operator T&& ();

    template<typename T, typename U = S, std::enable_if_t<
        std::is_same<T const&, U>::value, int> = 0>
    operator T const& () const;

    template<typename T, typename U = S&&, std::enable_if_t<
        std::is_same<T const&&, U>::value, int> = 0>
    operator T const&& () const;
};

struct some_type {};
struct other_type {};

auto test_call(some_type const&, other_type) -> std::false_type;
auto test_call(some_type&, other_type) -> std::true_type;

int main() {
    static_assert(decltype(test_call(probe<some_type&>{}, other_type{}))::value, "");
}

它可以在gcc和clang下工作,但不能在Visual Studio上编译,因为存在不明确的分辨率错误。哪个编译器错误,为什么?

GCC and Clang ,请 Visual studio

以下是MSVC输出:

source_file.cpp(31): error C2668: 'test_call': ambiguous call to overloaded function
source_file.cpp(28): note: could be 'std::true_type test_call(some_type &,other_type)'
source_file.cpp(27): note: or       'std::false_type test_call(const some_type &,other_type)'
source_file.cpp(31): note: while trying to match the argument list '(probe<some_type &>, other_type)'
source_file.cpp(31): error C2651: 'unknown-type': left of '::' must be a class, struct or union
source_file.cpp(31): error C2062: type 'unknown-type' unexpected

1 回复 | 直到 7 年前

1

xskxzr 7 年前

代码可以简化为 the following

#include <type_traits>

struct some_type {};

struct probe {
    template<typename T, std::enable_if_t<!std::is_const<T>::value, int> = 0>
    operator T& () const;
};

auto test_call(some_type const&) -> std::false_type;
auto test_call(some_type&) -> std::true_type;

int main() {
    static_assert(decltype(test_call(probe{}))::value, "");
}

5 6

一般来说,推导过程试图找到模板参数值,使推导出的值与a相同。但是,有四种情况允许差异:

……

T some_type 对于两个函数调用。然后根据[超过.ics.等级]。/ 3.3

probe -> some_type& -> some_type& probe -> some_type& -> const some_type& 所以没有歧义,GCC和Clang是对的。

顺便说一句,如果我们移除 std::enable_if_t<...> test_all

#include <type_traits>

struct some_type {};

struct probe {
    template<typename T>
    operator T& () const
    {
        static_assert(std::is_const_v<T>);
        static T t;
        return t;
    }
};

auto test_call(some_type const&) -> std::false_type;

int main() {
    test_call(probe{});
}

static_assert only under Clang .也就是说,Clang推断 一些类型 const some_type