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《C++并发编程》21页为什么说这里data传递的是副本，根据模板推导规则，传递不应该是...
source link: https://www.zhihu.com/question/449957754/answer/1786388445
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《C++并发编程》21页为什么说这里data传递的是副本，根据模板推导规则，传递不应该是左值引用吗？
《C++并发编程》原文代码 [图片] 2. 《Effective Modern C++》模板推导规则 [图片] 传递的data是左值，因此param形…
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事实上这段代码无法通过编译。因为你为一个函数调用而起一个线程，而这个函数入参传引用，那么无法保证入参的生命周期，很可能造成悬挂引用导致程序挂掉。
为了避免这种问题，标准库会进行静态检查，通过编译报错让你无法传引用。
static_assert(__is_invocable<typename decay<_Callable>::type,
typename decay<_Args>::type...>::value,
"std::thread arguments must be invocable after conversion to rvalues"
);
The arguments to the thread function are moved or copied by value. If a reference argument needs to be passed to the thread function, it has to be wrapped (e.g., with std::ref or std::cref).
Any return value from the function is ignored. If the function throws an exception, std::terminate is called. In order to pass return values or exceptions back to the calling thread, std::promise or std::async may be used.
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