c++ - 线程调用函数模板时的编译错误(对于默认参数)

Question

以下函数模板来自 Stroustrup 的书(“CPL”，第 4 版):

template<typename Arg1, typename Arg2, typename Arg3>
void write(Arg1 a1, Arg2 a2 = {}, Arg3 a3 = {});

template<typename Arg1, typename Arg2, typename Arg3>
void write(Arg1 a1, Arg2 a2, Arg3 a3)
{
   thread::id name = this_thread::get_id();

   coutm.lock();

   cout << "From thread " << name << " : "
        << a1 << ' ' << a2 << ' ' << a3
        << endl;

   coutm.unlock();
}

我正在尝试执行以下操作 here :
1) 调用此函数仅传递 2 个参数。第三个参数应该默认。
2) 函数模板调用直接传递给线程(而不是通过 lambda)。

int main()
{
   thread t (write<char, float>, 'a', 4.9);

   t.join();
}

但是，我得到以下编译错误:

main.cpp: In function 'int main()':
main.cpp:22:42: error: no matching function for call to 'std::thread::thread(<unresolved overloaded function type>, char, double)'

    thread t (write<char, float>, 'a', 4.9);
                                          ^
In file included from main.cpp:7:0:
/usr/local/include/c++/7.2.0/thread:118:7: note: candidate: template<class _Callable, class ... _Args> std::thread::thread(_Callable&&, _Args&& ...)

       thread(_Callable&& __f, _Args&&... __args)
       ^~~~~~
/usr/local/include/c++/7.2.0/thread:118:7: note:   template argument deduction/substitution failed:
main.cpp:22:42: note:   couldn't deduce template parameter '_Callable'

    thread t (write<char, float>, 'a', 4.9);
                                          ^
In file included from main.cpp:7:0:
/usr/local/include/c++/7.2.0/thread:113:5: note: candidate: std::thread::thread(std::thread&&)

     thread(thread&& __t) noexcept
     ^~~~~~

/usr/local/include/c++/7.2.0/thread:113:5: note:   candidate expects 1 argument, 3 provided
/usr/local/include/c++/7.2.0/thread:106:5: note: candidate: std::thread::thread()

     thread() noexcept = default;
     ^~~~~~

/usr/local/include/c++/7.2.0/thread:106:5: note:   candidate expects 0 arguments, 3 provided

对于默认参数，是否可以通过线程直接调用这样的函数模板？如果是这样怎么办？否则，必须将 lambda 传递给线程构造函数吗？

Answer 1

无法从默认(函数)参数推导出模板参数。您没有明确指定 Arg3，因此无法实例化 write。换句话说，您希望 a3 在线程对 write 的调用中是什么类型？