c++ - 没有前向声明的嵌套函数模板实例在 GCC 上编译，但不在 clang 上编译

Question

以下内容不能在 clang 中编译，但可以在 GCC ( godbolt ) 中编译:

template <typename K, typename V>
std::ostream& operator<< (std::ostream& o, const std::map<K, V>& map)
{
    const char* sep = "{";
    for (const auto& x : map)
    {
        o << sep << "{" << x.first << ", " << x.second << "}";
        sep = ", ";
    }
    return o << "}";
}

template <typename T>
std::ostream& operator<< (std::ostream& o, const std::vector<T>& vec)
{
    const char* sep = "{";
    for (const auto& x : vec)
    {
        o << sep << x;
        sep = ", ";
    }
    return o << "}";
}

// Usage

int main()
{
    std::map<int, std::vector<int>> t = {{1, {2, 3}}, {4, {5}}};
    std::cout << t << std::endl;
    return 0;
}

谁是对的？
顺便说一句，我知道它是 UB，但是将两个模板定义放在 namespace std 中也使代码在 clang 上编译。
代码借自 Is it possible to define operator<< for templated types?

Answer 1

叮当是对的。你应该搬家 operator<< std::vector的声明之前 operator<< std::map 的定义.
在 unqualified name lookup ,
(强调我的)

For a dependent name used in a template definition, the lookup is postponed until the template arguments are known, at which time ADL examines function declarations with external linkage (until C++11) that are visible from the template definition context as well as in the template instantiation context, while non-ADL lookup only examines function declarations with external linkage (until C++11) that are visible from the template definition context (in other words, adding a new function declaration after template definition does not make it visible except via ADL).