给定代码：

#include <iostream>

// this is my custom streamer which functions like ostream:
// MyStream() << 123 << "456" << 2.0 << ...;
class MyStream {
public:
    template <typename T>
    // It only supports streaming basic types,such as int/float/char/char* ...
    MyStream& operator<<(T&& t) {
        // std::cout is just an example, actually I stream `t` to a buffer
        std::cout << t << std::endl;
        return *this;
    }
};

struct Test {
    int x;
    float y;
};

// I want to stream struct Test:
// MyStream() << Test();
// It will be more useful if I can stream `Test` to iostream or stringstream.
// std::cout << Test();
// std::stringstream ss;
// ss << Test();
// So I implemented a template operator overload function

template <typename Stream>
Stream& operator<<(Stream& s, const Test& t) {
    return s << t.x << ", " << t.y;
}

int main() {
    const Test t = {1, 2.0};
    MyStream ms;
    // but this doesn't work as expected
    ms << t;

    Test t2 = {3, 4.0};
    ms << t2;
}

字符串
它生成编译错误：

source>:26:8: error: use of overloaded operator '<<' is ambiguous (with operand types 'MyStream' and 'const Test')
    ms << t;
    ~~ ^  ~
<source>:7:15: note: candidate function [with T = const Test &]
    MyStream& operator<<(T&& t) {
              ^
<source>:19:9: note: candidate function [with Stream = MyStream]
Stream& operator<<(Stream& s, const Test& t) {
        ^
1 error generated.

型
代码链接：https://godbolt.org/z/Gf51fjTMx
实际上，LogStream是第三方库fmt的 Package 器，我想让它像iostream一样工作。但它不支持自定义结构类型。因此，我添加了另一个重载函数模板Stream& operator<<(Stream& s, const Test& t)，以允许自定义结构的流。
问：
如果我不改变调用位置代码（main函数），如何使这两个函数重载中的任何一个更加专用化？