What's the error In named lambda inside class

Question

Hi

Refer code below :

It is working fine as per expectation.

I just created named lambda called otherlambda in main function and then called same.

When I am doing so inside class , I got the error. Yo get the error, just uncomment the code related to mylambda.

How to get these resolved ?
https://sololearn.com/compiler-playground/cNu9O1lFln2I/?ref=app

Answer 1

a placeholder type specifier (i.e. auto, decltype(auto) (since c++14) , a class template name subject to deduction (since C++17), a constrained placeholder (since C++20) ) cannot be used in a non-static data member declaration. (although it is allowed for static data members that are initialized in the class definition ).

in order to achieve you goal You could use < std::function > :

https://code.sololearn.com/cx13LnCU7d2y/?ref=app
https://www.sololearn.com/en/compiler-playground/cx13LnCU7d2y

Answer 2

don't use it?
The only way I got mylambda to work was if m_x was static const and mylambda was static constexpr. But that will make m_x useless...

lambda is just too restrictive. It have it's use case, but maybe using a regular function is better in this case.

Answer 3

Great and thank you MO ELomari

Answer 4

Bob_Li

just for practicing and discovering what we're allowed to do and when and what we don't .

Answer 5

MO ELomari
nice.
But doesn't this makes using a lambda not really easier than simply using a regular function? I don't see any advantage in doing it this way, other than extra complication.

void mylambda(){ std::cout << m_x << std::endl;};

Answer 6

MO ELomari
right. It is good to know.
C++ 's bag of tricks is a very big one indeed.😅

But sometimes, the saying "Just because we could, does not mean we should" is a good one to remember.

It is very easy to make the code more complicated than it needs to be.