[英]Passing a mutable lambda with unique_ptr into a const& std::function
I have got a dispatch function which executes a given lambda in a main thread. 我有一个调度函数,它在主线程中执行给定的lambda。 For the sake of this question, suppose it looks like the following:
为了这个问题,假设它看起来如下:
void dispatch(const std::function<void()>& fn) {
fn();
}
I need to load a new object in a new thread without interrupting the main thread. 我需要在新线程中加载一个新对象而不会中断主线程。 So I do the following: 1) start a new thread and create a new unique pointer inside the thread, 2) call
dispatch
and propagate the new unique pointer where it belongs. 所以我执行以下操作:1)启动一个新线程并在线程内创建一个新的唯一指针,2)调用
dispatch
并传播它所属的新唯一指针。
std::unique_ptr<std::string> foo; // nullptr
// do the loading in a new thread:
std::thread t([&](){
// in the new thread, load new value "Blah" and store it temporarily
auto bar = std::make_unique<std::string>("Blah");
dispatch([bar2 = std::move(bar), &foo]() mutable {
foo = std::move(bar2); // propagate the loaded value to foo
});
});
t.join(); // for the sake of this example
std::cout << "foo = " << *foo << std::endl; // this should say: foo = Blah
Run example online: http://cpp.sh/5zjvm 在线运行示例: http : //cpp.sh/5zjvm
This code does not compile because the inner lambda in dispatch
is mutable
and so does not fit into dispatch(const std::function<void()>& fn)
which requires a const&
. 此代码无法编译,因为
dispatch
的内部lambda是mutable
,因此不适合需要const&
dispatch(const std::function<void()>& fn)
。
The lambda, however, needs to be mutable
because it needs to call std::move
on the unique pointers. 然而,lambda需要是
mutable
因为它需要在唯一指针上调用std::move
。
This code could be fixed for example by changing dispatch
to: 可以修改此代码,例如将
dispatch
更改为:
template <typename Fn>
void dispatch(Fn fn) {
fn();
}
Unfortunately, the dispatch
function is an API of a library and I cannot change it. 不幸的是,
dispatch
功能是库的API,我不能改变它。
Is there a way out of this problem without getting rid of unique pointers? 如果没有摆脱独特的指针,有没有办法摆脱这个问题?
No, that isn't your problem. 不,那不是你的问题。
Your problem is that your lambda cannot be copied, as it has a unique ptr captured by value in it. 你的问题是你的lambda无法复制,因为它有一个由值中捕获的唯一ptr。
std::function<Sig>
type erases down to std::function<Sig>
类型删除到
Invoke with Sig
用
Sig
调用
Destroy 破坏
Copy (and sometimes move) 复制(有时移动)
Cast-back-to-original-type 铸造回至原来的型
Your lambda cannot be copied, so cannot be stored in a std::function
. 您的lambda无法复制,因此无法存储在
std::function
。
The lazy-coder's solution is: 懒惰编码器的解决方案是:
dispatch([bar2 = std::make_shared<decltype(bar)>(std::move(bar)), &foo]() mutable {
foo = std::move(*bar2);
});
where we shove the non-copyable state into a shared_ptr
. 我们将不可复制的状态推送到
shared_ptr
。
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