Capture arguments are copied from rvalue lambda? [duplicate]

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• How to create an std::function from a move-capturing lambda expression? 3 answers

Are captured arguments copied during the conversion of lambdas to std::function?
I need to convert a lambda that captures a non-copyable type to std::function.
So I passed a lambda to std::function as an rvalue, but an error occurred.

// Foo is non-copyable.
auto a = [f = Foo()]{ };
std::function<void()> b = std::move(a) // error, calls deleted Foo::Foo(const Foo&);