在可变参数模板中为每个压缩参数执行函数

Question

I noticed the following line in the open-source project FeatherKit :

int _[] = { (SubscribeToType<MessageTypes>( bus, receiver, desubscribers, unsubscribe ), 0)... };

With the following context:

template<class... MessageTypes>
void Subscribe( MessageBus& bus, MessageReceiver<MessageTypes...>& receiver, bool unsubscribe ) {
    std::vector<std::function<void()>> desubscribers;
    int _[] = { (SubscribeToType<MessageTypes>( bus, receiver, desubscribers, unsubscribe ), 0)... };
    (void) _;
    receiver.desubscribers = desubscribers;
}

It's obviously executing the function SubscribeToType for each parameter in the variadic template.

My question is twofold:

1. How, exactly, does the line work? How come parameter unpacking is allowing that function to execute for each parameter in the variadic template?

2. I am very certain this line could be replaced by a lambda. How could you replace the line with a lambda expression?

I've contacted the original author of FeatherKit, but he wasn't able to answer my question at that time.

Answer 1

1. How, exactly, does the line work? How come parameter unpacking is allowing that function to execute for each parameter in the variadic template?

A parameter pack expansion is some pattern involving a parameter pack followed by ...

So expr(T)... is a pack expansion with expr(T) as its pattern, and it expands to expr(T0), expr(T1), expr(T2), ..., expr(TN) for each Ti in the parameter pack.

A pack expansion can only be used in certain contexts, such as argument lists, or initializer lists, so in this case the results of each sub-expression is being used to form an initializer list for the array int _[] . The array is unused and only exists so that its initializer can be used as the context in which to do the pack expansion. Each sub-expression is of the form (SubscribeToType<Ti>(blah, blah), 0) which means the result of the function call is discarded and the expression has the value 0. This is a bit of a kluge to allow the pack expansion to produce a braced-init-list containing N integers, because that's what's needed to initialize the array.

2. I am very certain this line could be replaced by a lambda. How could you replace the line with a lambda expression?

Why would you want to?

It could, but you'd need a very similar pack expansion in the lambda, so it wouldn't simplify anything.