为何为接收std::string引用的talk函数创建std::thread会编译报错？

我写了一个talk函数，能逐字符延迟输出传入的字符串，单独调用完全正常，但用std::thread创建线程执行时出现大量编译错误。而无参数的hi函数创建线程却能正常运行。

talk函数代码 #

void talk(std::string& string)
{
    for (volatile int i{0}; i < string.length(); ++i)
    {
        if (string[i] != '\n')
        {
            if (i > 0 && string[i - 1] == '\n') std::cout << '\n';
            std::cout << string[i] << std::flush;
            std::this_thread::sleep_for(std::chrono::milliseconds(100));
        } else std::this_thread::sleep_for(std::chrono::milliseconds(200));
    }
    std::this_thread::sleep_for(std::chrono::milliseconds(300));
}

无法编译的完整代码 #

#include <iostream>
#include <thread>
#include <chrono>

void talk(std::string& string)
{
    for (volatile int i{0}; i < string.length(); ++i)
    {
        if (string[i] != '\n')
        {
            if (i > 0 && string[i - 1] == '\n') std::cout << '\n';
            std::cout << string[i] << std::flush;
            std::this_thread::sleep_for(std::chrono::milliseconds(100));
        } else std::this_thread::sleep_for(std::chrono::milliseconds(200));
    }
    std::this_thread::sleep_for(std::chrono::milliseconds(300));
}

int main()
{
    std::string message{"Hello, how are you doing?"};
    std::thread talkThread{talk, message};
    talkThread.join();
    return 0;
}

正常运行的无参数函数代码 #

#include <iostream>
#include <thread>

void hi()
{
    std::cout << "Hi.";
}

int main()
{
    std::thread hiThread{hi};
    hiThread.join();
    return 0;
}

编译报错信息 #

In file included from /usr/include/c++/11/thread:43,
                 from main.cpp:2:
/usr/include/c++/11/bits/std_thread.h: In instantiation of ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(std::__cxx11::basic_string<char>&); _Args = {std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&}; <template-parameter-1-3> = void]’:
main.cpp:27:41:   required from here
/usr/include/c++/11/bits/std_thread.h:130:72: error: static assertion failed: std::thread arguments must be invocable after conversion to rvalues
  130 |                                       typename decay<_Args>::type...>::value,
      |                                                                        ^~~~~
/usr/include/c++/11/bits/std_thread.h:130:72: note: ‘std::integral_constant::value’ evaluates to false
/usr/include/c++/11/bits/std_thread.h: In instantiation of ‘struct std::thread::_Invoker<std::tuple<void (*)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > > >’:
/usr/include/c++/11/bits/std_thread.h:203:13:   required from ‘struct std::thread::_State_impl<std::thread::_Invoker<std::tuple<void (*)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > > > >’:
/usr/include/c++/11/bits/std_thread.h:143:29:   required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(std::__cxx11::basic_string<char>&); _Args = {std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&}; <template-parameter-1-3> = void]’:
main.cpp:27:41:   required from here
/usr/include/c++/11/bits/std_thread.h:258:11: error: no type named ‘type’ in ‘struct std::thread::_Invoker, std::allocator >&), std::__cxx11::basic_string, std::allocator > > >::__result, std::allocator >&), std::__cxx11::basic_string, std::allocator > > >’
  258 |           _M_invoke(_Index_tuple<_Ind...>)
      |           ^~~~~~~~~
/usr/include/c++/11/bits/std_thread.h:262:9: error: no type named ‘type’ in ‘struct std::thread::_Invoker, std::allocator >&), std::__cxx11::basic_string, std::allocator > > >::__result, std::allocator >&), std::__cxx11::basic_string, std::allocator > > >’
  262 |         operator()()
      |         ^~~~~~~~

错误原因 #

核心问题在于std::thread的构造函数会对传入的参数执行类型衰减（decay），也就是把左值转换成右值。而talk函数的参数是std::string&（非const左值引用），无法绑定到衰减后的右值std::string，导致编译器认为参数无法被调用，触发静态断言失败。

解决方案 #

方案1：用std::ref传递引用 #

在创建线程时，用std::ref包裹参数，让std::thread内部保持对原变量的引用：

int main()
{
    std::string message{"Hello, how are you doing?"};
    std::thread talkThread{talk, std::ref(message)};
    talkThread.join();
    return 0;
}

方案2：修改函数参数为const引用 #

因为talk函数只是读取字符串，没有修改操作，所以可以把参数改成const std::string&，这样既可以接受左值，也能兼容std::thread的参数衰减：

void talk(const std::string& string)
{
    // 函数内容不变
}

// main函数无需修改，直接传message即可
int main()
{
    std::string message{"Hello, how are you doing?"};
    std::thread talkThread{talk, message};
    talkThread.join();
    return 0;
}

方案3：传递指针 #

如果确实需要非const引用（比如函数要修改字符串），也可以传递指针，避免引用绑定问题：

void talk(std::string* string)
{
    if (!string) return; // 空指针检查
    for (volatile int i{0}; i < string->length(); ++i)
    {
        if ((*string)[i] != '\n')
        {
            if (i > 0 && (*string)[i - 1] == '\n') std::cout << '\n';
            std::cout << (*string)[i] << std::flush;
            std::this_thread::sleep_for(std::chrono::milliseconds(100));
        } else std::this_thread::sleep_for(std::chrono::milliseconds(200));
    }
    std::this_thread::sleep_for(std::chrono::milliseconds(300));
}

int main()
{
    std::string message{"Hello, how are you doing?"};
    std::thread talkThread{talk, &message};
    talkThread.join();
    return 0;
}

内容的提问来源于stack exchange，提问作者Jerry