如何定义模板化外部结构的内部结构的 ostream 函数？

Question

我怎样才能让它发挥作用？我想在这里将friend ostream 函数的定义和声明分开。

#include <bits/stdc++.h>
using namespace std;

template <class T>
struct Outer {
    struct Inner {
        T x;
    };
    friend std::ostream& operator<<(std::ostream&, const Inner&);
};

template <class T>
std::ostream& operator<<(std::ostream& os, const Outer<T>::Inner& i) {
    os << "x! " << i.x;
    return os;
}

int main() {
    const Outer<int>::Inner s = {42};
    std::cout << s << "\n";
    return 0;
}

按照错误的建议添加类型名也不起作用

template <class T>
std::ostream& operator<<(std::ostream& os, const typename Outer<T>::Inner& i) {
    os << "x! " << i.x;
    return os;
}

warning: friend declaration 'std::ostream& operator<<(std::ostream&, const Outer<T>::Inner&)' declares a non-template function [-Wnon-template-friend]
   10 |     friend std::ostream& operator<<(std::ostream&, const Inner&);
      |                                                                ^
<source>:10:64: note: (if this is not what you intended, make sure the function template has already been declared and add '<>' after the function name here)
/opt/compiler-explorer/gcc-13.2.0/bin/../lib/gcc/x86_64-linux-gnu/13.2.0/../../../../x86_64-linux-gnu/bin/ld: /tmp/ccBSBhtj.o: in function `main':
<source>:24: undefined reference to `operator<<(std::ostream&, Outer<int>::Inner const&)'
collect2: error: ld returned 1 exit status

Answer 1

与：

template <class T>
std::ostream& operator<<(std::ostream& os, const typename Outer<T>::Inner& i)

T

不可推论。

将声明与定义分开并不明显。

会更简单

template <class T>
struct Outer {
    struct Inner {
        T x;

        friend std::ostream& operator<<(std::ostream&, const Inner&) {
            return os << "x! " << i.x;
        }
    };
};