呼叫运营商<<临时

考虑以下代码(godbolt(：

#include <fstream>
#include <string>
#include <filesystem>
namespace fs = std::filesystem;
using namespace std::string_literals;
template<class CharT, class Traits>
void test(std::basic_ostream<CharT, Traits> &s) {
s << "Other testn"s;
}
struct my_struct {
std::string s = "My Structn"s;
};
template<class CharT, class Traits>
std::basic_ostream<CharT, Traits> &operator<<(std::basic_ostream<CharT, Traits> &stream,
const my_struct &s) {
stream << s.s;
return stream;
}
int main() {
fs::path file("test.txt");
const my_struct s;
// A
operator<<(std::ofstream(file, std::ios_base::app), "Other test!n"s);   // Runs fine
std::ofstream(file, std::ios_base::app) << "Das ist ein Test!n"s;       // Runs fine
// B
operator<<(std::ofstream(file, std::ios_base::app), s);                  // Runs fine
std::ofstream(file, std::ios_base::app) << s;                            // Runs fine
// C
// test(std::ofstream(file, std::ios_base::app));                        // Does not compile
}

代码部分A和B编译得很好，而C则不然。很明显，test不能取右值，因为参数是非const引用。

但是A和B不应该是一样的吗？STL的operator<<和我自己的都把左手边作为非const的参考。那么，为什么他们可以使用临时对象呢？

编译器：

clang version 10.0.0 
Target: x86_64-pc-linux-gnu
Thread model: posix
InstalledDir: /usr/bin
Found candidate GCC installation: /usr/bin/../lib/gcc/x86_64-pc-linux-gnu/10.1.0
Found candidate GCC installation: /usr/bin/../lib64/gcc/x86_64-pc-linux-gnu/10.1.0
Found candidate GCC installation: /usr/lib/gcc/x86_64-pc-linux-gnu/10.1.0
Found candidate GCC installation: /usr/lib64/gcc/x86_64-pc-linux-gnu/10.1.0
Selected GCC installation: /usr/bin/../lib64/gcc/x86_64-pc-linux-gnu/10.1.0
Candidate multilib: .;@m64
Candidate multilib: 32;@m32
Selected multilib: .;@m64

编辑：这个问题的第一个版本有一个错误，导致了自己的operator<<segfault。