CODEWITHSUNDEEP
c++c++11lambdacopy-constructor
ksl
ksl

Reputation: 4709

Lambda capture by value and non copyable object

I'm trying to understand the implications of the lambda expression in the code snippet below.

The lambda expression captures variables by value rather than by reference otherwise the local variable message is destroyed when foo exits.

What I don't understand is the capture of m_impl. How is it captured by value if the copy ctor of Impl is deleted? Please can someone enlighten me?

void Foo::foo(std::shared_ptr<std::string> message)
{
    m_impl->m_thread.send([=] { m_impl->handleMessage(message); });
}

handleMessage is declared as:

void handleMessage(std::shared_ptr<std::string> message)

and m_impl as:

std::unique_ptr<Impl> m_impl;

Impl has its copy constructor and assignment operator deleted.

Upvotes: 1

Views: 925

Answers (1)

Mike Seymour
Mike Seymour

Reputation: 254531

The things that can be captured are:

  • local variables, by value or reference;
  • in a class member function, the this pointer, by value.

Member variables are not captured, although capturing this effectively captures them by reference. By specifying a capture default, this will be captured if you refer to any class member within the lambda body, allowing access to that member.

Presumably, m_impl is a member variable, so that's what's happening here.

Upvotes: 7

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