使用 auto 推导的 lambda 成员函数模板

Question

In the code below, events is an stl container built from template policies where the underlying type is composed from types in inbound and outbound.

auto events = process::run<inbound, outbound>(args);

After processing events, I need to do some calculations based on runtime parameters. Here's an example:

cool::transform(events.deltas(), [](const auto &current) { return current.method(); });

There are several methods we may want to use, ie we may call another method which may or may not return the same underlying type.

cool::transform(events.deltas(), [](const auto &current) { return current.other(); });

Is there a way to template this so I can declare:

cooler::transform<method>(events.deltas())

Bonus question: Can we use templates to only enable the code if the method actually exists on the underlying type?

Answer 1

Is there a way to template this so I can declare:

Not really - there is no gotcha-free way of passing "a method" as a template parameter. You will have issues with overloaded/template member functions. The lambda solution you currently have is the best one - I suggest shortening current to c and using auto&& in order to reduce the boilerplate:

cool::transform(events.deltas(), [](auto&& c){ return c.method(); });

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Can we use templates to only enable the code if the method actually exists on the underlying type?

Yes, you can use the detection idiom to check whether or not an expression is valid. I wrote an article that covers it and other more powerful/terser techniques: "checking expression validity in-place with C++17" .