How to call a function from a pointer to that function?

Question

I'm struggling with maybe a very simple hurdle but can't seem to understand how that works.

In my C++ project I want to provide lazy loaded classes. For this I thought of instantiating some fields of the type

std::shared_ptr<std::function<std::shared_ptr<MyClass>>> myClassLazyLoader;

At this point I'm not even sure anymore that is a valid approach, so please correct me if I'm completely off track.

I instantiate the field like this:

myClassLazyLoader = [] { return std::make_shared<MyClass>(); }

To invoke the function, I tried

myClassLazyLoader()

after which the compiler told me, as expected, that std::shared_ptr<std_function<... does not provide a call operator.

Naturally, I tried

(*myClassLazyLoader)()

but then it told me that std::function<... also does not provide a call operator which doesn't make any sense to me.

What am I doing wrong?

Thanks much in advance!

Can you extract and provide a minimal reproducible example? Also, the first attempt is obviously flawed, but the second one not so, so please include the exact and full error it produces. — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Nov 8, 2021 at 16:45
You have a typo in template definition, missing () - it should be std::function<std::shared_ptr<MyClass>()>. — Yksisarvinen
– Yksisarvinen, Commented Nov 8, 2021 at 16:46

rombas · Accepted Answer · 2021-11-08 16:48:17Z

1

The moment I posted this I figured it out. The type should be

std::shared_ptr<std::function<std::shared_ptr<MyClass>()>>

answered Nov 8, 2021 at 16:48

rombas

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