没有前向声明的嵌套函数模板实例化在 GCC 上编译但在 clang 上不编译

Question

以下代码不能在 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;
}

谁说得对？

顺便说一句，我知道它是 UB，但是将两个模板定义放在 namespace std 中也会使代码在 clang 上编译。

代码借自Is it possible to define operator<< for templated types?

Answer 1

铿锵是对的。您应该在 operator<< 对 std::map 的定义之前移动 operator<< 对 std::vector 的声明。

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).

没有前向声明的嵌套函数模板实例化在 GCC 上编译但在 clang 上不编译

Nested function template instantiation without forward declaration compiles on GCC but not on clang

c++

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